% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% The Project Gutenberg EBook of Solid Geometry with Problems and         %
% Applications (Revised edition), by H. E. Slaught and N. J. Lennes       %
%                                                                         %
% This eBook is for the use of anyone anywhere at no cost and with        %
% almost no restrictions whatsoever.  You may copy it, give it away or    %
% re-use it under the terms of the Project Gutenberg License included     %
% with this eBook or online at www.gutenberg.org                          %
%                                                                         %
%                                                                         %
% Title: Solid Geometry with Problems and Applications (Revised edition)  %
%                                                                         %
% Author: H. E. Slaught                                                   %
%         N. J. Lennes                                                    %
%                                                                         %
% Release Date: August 26, 2009 [EBook #29807]                            %
%                                                                         %
% Language: English                                                       %
%                                                                         %
% Character set encoding: ISO-8859-1                                      %
%                                                                         %
% *** START OF THIS PROJECT GUTENBERG EBOOK SOLID GEOMETRY ***            %
%                                                                         %
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%%%%%%%%%%%%%%%%%%%%%%%%%% BEGIN DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%%%

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The Project Gutenberg EBook of Solid Geometry with Problems and
Applications (Revised edition), by H. E. Slaught and N. J. Lennes

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Solid Geometry with Problems and Applications (Revised edition)

Author: H. E. Slaught
        N. J. Lennes

Release Date: August 26, 2009 [EBook #29807]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK SOLID GEOMETRY ***
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\begin{BioSketch}{Cavalieri}{Bonaventura Cavalieri}
(1598--1647) was one of the most influential
mathematicians of his time. He was chiefly noted for his invention
of the so-called ``Principle of Indivisibles'' by which he derived areas
and volumes. See pages \Pgref{134} and~\Pgref{202}.
\end{BioSketch}

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{\MyHuge SOLID GEOMETRY} \\[24pt]

{\footnotesize WITH} \\[24pt]

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{\large \textit{REVISED EDITION}} \\[36pt]

BY \\[18pt]

{\Large H.~E.~SLAUGHT, \textsc{Ph.D}., \textsc{Sc.D}.} \\[8pt]

{\scriptsize PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CHICAGO} \\[12pt]

AND \\[18pt]

{\Large N. J. LENNES, \textsc{Ph.D}.} \\[8pt]

{\scriptsize PROFESSOR OF MATHEMATICS IN THE UNIVERSITY \\[-2pt]
OF MONTANA} \\[210pt]

{\Large \textsc{ALLYN and BACON}} \\[12pt]

{\large\textgoth{Boston}\qquad\qquad \textgoth{New York}\qquad\qquad \textgoth{Chicago}}
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COPYRIGHT, 1919, BY \\[8pt]

H.~E.~SLAUGHT AND N.~J.~LENNES \\[36pt]

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%%%% Credits and transcriber's note %%%%
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Produced by Peter Vachuska, Andrew D. Hwang, Chuck Greif
and the Online Distributed Proofreading Team at
http://www.pgdp.net
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%The original book was published in 1919 by Allyn and Bacon.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PREFACE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\section*{\centering PREFACE}

In re-writing the Solid Geometry the authors have consistently
carried out the distinctive features described in the
preface of the Plane Geometry. Mention is here made only
of certain matters which are particularly emphasized in the
Solid Geometry.

Owing to the greater maturity of the pupils it has been possible
to make the logical structure of the Solid Geometry more
prominent than in the Plane Geometry. The axioms are
stated and applied at the precise points where they are to be
used. Theorems are no longer quoted in the proofs but are
only referred to by paragraph numbers; while with increasing
frequency the student is left to his own devices in supplying
the reasons and even in filling in the logical steps of the argument.
For convenience of reference the axioms and theorems
of plane geometry which are used in the Solid Geometry are
collected in the Introduction.

In order to put the essential principles of solid geometry,
together with a reasonable number of applications, within
% [** PP: Hard-coded page quantity]
limited bounds (156 pages), certain topics have been placed in
an Appendix. This was done in order to provide a minimum
course in convenient form for class use and not because these
topics, Similarity of Solids and Applications of Projection, are
regarded as of minor importance. In fact, some of the examples
under these topics are among the most interesting and
concrete in the text. For example, see pages
\Pgref{170}--\Pgref{172}, \Pgref{177a}--\Pgref{177b},
\Pgref{183}--\Pgref{184}.

The exercises in the main body of the text are carefully
graded as to difficulty and are not too numerous to be easily
performed. The concepts of three-dimensional space are made
%% -----File: 005.png---Folio xx-------
clear and vivid by many simple illustrations and questions
under the suggestive headings ``Sight Work.'' This plan of
giving many and varied simple exercises, so effective in the
Plane Geometry, is still more valuable in the Solid Geometry
where the visualizing of space relations is difficult for many
pupils.

The treatment of incommensurables throughout the body of
this text, both Plane and Solid, is believed to be sane and
sensible. In each case, a frank assumption is made as to the
existence of the concept in question (length of a curve, area of
a surface, volume of a solid) and of its realization for all practical
purposes by the approximation process. Then, for theoretical
completeness, rigorous proofs of these theorems are
given in \hyperref[appendix:III]{Appendix~III}, where the theory of limits is presented
in far simpler terminology than is found in current text-books
and in such a way as to leave nothing to be unlearned or compromised
in later mathematical work.

Acknowledgment is due to Professor David Eugene Smith
for the use of portraits from his collection of portraits of
famous mathematicians.

\null\hfill\begin{tabular}{l}
H.~E.~SLAUGHT \\
N.~J.~LENNES
\end{tabular}\hspace*{1em}

\noindent{\setlength{\tabcolsep}{0pt}\footnotesize%
\begin{tabular}{c}
\textsc{Chicago} and \textsc{Missoula}, \\
May, 1919.
\end{tabular}}


\NewFolio %% -----File: 006.png---Folio xx-------

%%%%%%%%%%%%%%%%%%%%%%%%% TABLE of CONTENTS %%%%%%%%%%%%%%%%%%%%%%%%%

\iffalse %%%% OCRed Table of Contents %%%%
CONTENTS

PAGE

INTRODUCTION .......... 1

Space Concepts .......... 1

Axioms and Theorems from Plane Geometry .... 5


BOOK I. Properties OF THE Plane ...... 9

Perpendicular Planes and Lines ....... 10

Parallel Planes and Lines ........ 20

Dihedral Angles .......... 28

Constructions of Planes and Lines ...... 35

Polyhedral Angles ......... 40

BOOK II. Regular Polyhedrons ...... 49

Construction of Regular Polyhedrons ..... 52

BOOK III. Prisms AND Cylinders ...... 53

Properties of Prisms ......... 54

Properties of Cylinders ........ 69

BOOK IV. Pyramids AND Cones ....... 79

Properties of Pyramids ........ 80

Properties of Cones ......... 92

BOOK V. The Sphere ......... 105

Spherical Angles and Triangles ....... 117

Area of the Sphere ......... 136

Volume of the Sphere ........ 141

%% -----File: 007.png---Folio xx-------

APPENDIX TO SOLID GEOMETRY

PAGE

I. Similar Solids             157

II. Applications of Projection             173

III. Theory of Limits              185

INDEX           207

PORTRAITS AND BIOGRAPHICAL SKETCHES

Cavalieri          Frontispiece

FACING PAGE

Thales              48

Archimedes          104

Legendre                156
\fi % End of OCRed Table of Contents


\phantomsection
\pdfbookmark[0]{Table of Contents}{Table of Contents}

\tableofcontents

\thispagestyle{empty}
\fancyhead[CO]{\textit{CONTENTS}}


\NewFolio %% -----File: 008.png---Folio 0-------

\null\vfill
\pagestyle{empty}
\begin{center}
  \LARGE SOLID GEOMETRY
\end{center}
\vfill

%% -----File: 009.png---Folio 1-------

%%%%%%%%%%%%%%%%%%%%%%%%%%%% MAIN MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mainmatter
\pagestyle{fancy}
\fancyhead{}
\fancyfoot{}

\renewcommand{\NewFolio}{}

\Chapter{\texorpdfstring{INTRODUCTION}{Introduction}}{}

\phantomsection
\SecToc{Space Concepts}

\Paragraph{1.}{Two-Dimensional Figures.} In plane geometry each figure
is restricted so that all of its parts lie in the same plane.
Such figures are called \emph{two-dimensional figures}.

\begin{Remark}
A figure, all parts of which lie in one straight line, is a \emph{one-dimensional
figure}, while a point is of zero dimensions.
\end{Remark}

\Paragraph{2.}{Three-Dimensional Figures.} A figure, not all parts of
which lie in the same plane, is a \emph{three-dimensional figure}.
\index{Figures@\textbf{Figures}!in space}%{\Pnum{2}}%

\begin{Remark}
Thus, a figure consisting of a plane and a line not in the plane is a
three-dimensional figure because the whole figure does not lie in one
plane.
\end{Remark}

\Paragraph{3.}{Solid Geometry} treats of the properties of three-dim\-en\-sion\-al
figures.
\index{Geometry@\textbf{Geometry}}%{\Pnum{3}}%

\Paragraph{4.}{Representation of a Plane.} While a plane is endless in
extent in all its directions, it is represented by a parallelogram,
or some other limited plane figure.

%[Illustration]
\begin{center}
\Input{009ab}{\textwidth}
\end{center}

A plane is designated by a single letter in it, by two letters
at opposite corners of the parallelogram representing it, or by
any three letters in it but not in the same straight line.

\begin{Remark}
Thus, we say \emph{the plane~$M$, the plane~$PQ$, or the plane~$ABC$}.
\end{Remark}

\NewFolio %% -----File: 010.png---Folio 2-------

\Paragraph{5.}{Figures in Plane and Solid Geometry.} In describing a
\index{Figures@\textbf{Figures}!in space}%{\Pnum{5}}%
figure  in  plane  geometry, it is assumed, usually without
special mention, that all parts of the figure lie in the same
plane, while in solid geometry it is assumed that the whole
figure need not lie in any one plane.

\begin{Remark}
Thus, in plane geometry we have the theorem:

``\emph{Through a fixed point on a line one and only one perpendicular can
be drawn to the line.}''
\end{Remark}

%\begin{wrapfigure}[3]{r}{1.5in}
%  \vspace*{-0.5in}
%  \Input{010a}{1.25in}
%\end{wrapfigure}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent If all parts of the figure are not required to lie in one plane,
the theorem just quoted is far from
true. As can be seen from the figure,
an unlimited number of lines can be
drawn perpendicular to a line at a
point in it.
\end{minipage}%
%[Illustration]
\hfill\smash{\Input{010a}{1.75in}}

\begin{Remark}
Thus, all the spokes of a wheel may be
perpendicular to the axle.
\end{Remark}


\Paragraph{6.}{Loci in Plane and Solid Geometry.} In plane geometry,
``the locus of all points at a given distance from a given
point'' is a circle, while in solid geometry this locus is a
sphere.

%[Illustrations]
\begin{center}
\Input{010bcd}{\textwidth}
\end{center}

In plane geometry, ``the locus of all points at a given distance
from a given line'' consists of two lines, each parallel
%
%[Illustration]
%
to the given line and at the given distance from it, while in
solid geometry this locus is a cylindrical surface whose radius
is the given distance.
\begin{center}
\Input{010e}{2in,height=0.875in}\qquad\qquad
\Input{010f}{2in}
\end{center}

\NewFolio %% -----File: 011.png---Folio 3-------

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{7.}{Parallel Lines. Skew Lines.} In plane geometry, two
lines which do not meet are parallel, while in solid geometry,
two lines which do not meet need not be parallel. That is,
they may not be in the same plane. Lines which are not
parallel and do not meet are called \emph{skew} lines.
\end{minipage}%
%[Illustration]
\hfill\Input{011a}{1.5in}

\begin{Remark}
In solid geometry, as in plane
geometry, the definition of parallel
lines implies that the lines lie in the
same plane. That is, if two lines
are parallel, there is always some
plane in which both lie. Thus, in
the figure, $l_1$~and $l_2$ are parallel, as
are also $l_1$~and $l_3$, while $l_3$~and $l_4$ are
skew.
\end{Remark}

\begin{Sightwork}

\index{Loci@\textbf{Loci problems}|(}%
\item[] \textit{Note}. In exercises 1--4 give the required loci for both plane and solid
geometry. No proofs are required.

\item[1.] The locus of all points six inches distant from a given point.

\item[2.] The locus of all points ten inches distant from a given point.

\item[3.] The locus of all points at a perpendicular distance of four inches
from a given straight line.

\item[4.] The locus of all points at a perpendicular distance of nine inches
from a given straight line.

\item[5.] Find the locus of all points one foot from a given plane. Is this
a problem in plane or in solid geometry?

\item[6.] Find the locus of all points equidistant from two parallel lines and
in the same plane with them. Is this a problem in plane or in solid
geometry?

\item[7.] Find the locus of all points equidistant from two given parallel
planes. Is this a problem in plane or in solid geometry?
\index{Loci@\textbf{Loci problems}|)}%

\item[8.] The side walls of your schoolroom meet each other in four vertical
lines. Are any two of these parallel? Are any three of them
parallel? Do any three of them lie in the same plane?

\item[9.] The side walls of your schoolroom meet the floor and the ceiling
in straight lines. Which of these lines are parallel to each other? Do
any of these lines lie in the same plane?

\end{Sightwork}

\NewFolio %% -----File: 012.png---Folio 4-------

\Paragraph{8.}{Representation of Solid Figures on a Plane Surface.} To represent
\index{Figures@\textbf{Figures}!in space}%{\Pnum{8}}%
a figure on a plane surface when at least part of the
figure does not lie in that surface requires special devices.

\begin{Remark}
%[Illustration]
\begin{wrapfigure}[6]{r}{2in}
\vspace*{-0.125in}
\Input{012a}{2in}
\end{wrapfigure}
Thus, in the parallelogram
$ABCD$ used to represent a plane,
the edges $AB$~and $BC$ are made
heavier than the other two. This
indicates that the lower and right-hand
sides are nearer the observer
than the other edges. Hence, the
plane represented does not lie in the plane of the
paper, but the lower part of it stands out toward
the observer.

%[Illustration]
\begin{wrapfigure}[7]{r}{2.5in}
\vspace*{-0.0625in}
% [** PP: Relocate second image from next paragraph]
\Input{012b}{1.25in}\hfill\Input{012c}{1.25in}
\end{wrapfigure}
The figure $ABCD$ represents a triangular pyramid.
The corner marked~$B$ is nearest the observer
and this is indicated by the heavy lines. The triangle
$ACD$ lies behind the pyramid and is thus farther
from the observer. The line~$AC$ is dotted to
indicate that it is seen through the figure.

%[Illustration]
In the closed box~$AG$, the lines $AD$, $DC$, and~$DH$ lie
behind the figure and are dotted, while the others are
in full view and are solid. If the box were open at the
top, part of the line~$DH$ would be in full view and
would be represented by a solid line.
\end{Remark}

\Paragraph{9.}{Representation of Lines.} The following
\index{Figures@\textbf{Figures}!in space}%{\Pnum{9}}%
plan for representing lines is generally adhered
to in this book:

\begin{Remark}

\begin{wrapfigure}[6]{l}{1.5in}
\vspace*{-0.125in}
\Input{012d}{1.5in}
\end{wrapfigure}

(1) A line of the main figure which is not obscured by any other part
of the figure is represented by a solid line.

(2) An auxiliary line, which is drawn incidentally
in making a proof or constructing a
figure, is marked in long dashes if it is in full view.

(3) Any line whatever which is behind a part
of the figure is marked in short dashes or dots, or
sometimes is not shown at all.

(4) Where a figure is shaded it is usually regarded as opaque and
the lines behind it cannot be seen at all.

(5) In some cases a shaded surface is regarded as translucent and the
lines behind it are seen dimly. Such lines are marked in short dashes.

\end{Remark}

\NewFolio %% -----File: 013.png---Folio 5-------

The following Axioms and Theorems from plane geometry
are referred to in the solid geometry. The special axioms of
solid geometry will be given as they arise in the text.

\SecToc{Axioms and Theorems from Plane Geometry}

\index{Theorems@\textbf{Theorems of Plane Geometry}|(}%{\Pnum{9}}%

\begin{Axioms}

\index{Axioms@\textbf{Axioms}|(}%{\Pnum{10}}%
\item[10.] Things equal to the same things are equal to each other.

\item[11.] If equals are added to equals, the sums are equal.

\item[12.] If equals are subtracted from equals, the remainders are
equal.

\item[13.] If equals are multiplied by equals, the products are equal.

\item[14.] If equals are divided by equals, the quotients are equal.

\item[15.] If equals are added to unequals, the sums are unequal and
in the same order.

\item[16.] If unequals are added to unequals, in the same order, then
the sums are unequal and in that order.

\item[17.] If equals are subtracted from unequals, the remainders are
unequal and in the same order.

\item[18.] If unequals are subtracted from equals, the remainders are
unequal and in the opposite order.

\item[19.] If $a$ is less than $b$~and $b$ less than~$c$, then $a$ is less than~$c$.

\item[20.] If $a$~and $b$ are quantities of the same kind, then either
$a > b$, or $a = b$, or $a < b$.
\Unit{20.}%

\item[21.] Through a point not on a given line only one straight line
can be drawn parallel to that line.
\Unit{21.}%

\item[22.] A straight line-segment is the shortest distance between
two points.

\item[23.] Corresponding parts of equal figures are equal.
\Unit{23.}%
\index{Axioms@\textbf{Axioms}|)}%{\Pnum{23}}%
\end{Axioms}


\begin{Theorems}

\item[24.] If two lines intersect, the vertical angles are equal.
\Unit{24.}%

\item[25.] Two triangles are equal if two sides and the included
angle of one are equal respectively to two sides and the included
angle of the other.

\NewFolio %% -----File: 014.png---Folio 6-------

\item[26.] Two triangles are equal if two angles and the included side
of one are equal respectively to two angles and the included side
of the other.

\item[27.] Two triangles are equal if three sides of one are equal
respectively to three sides of the other.
\Unit{27.}%

\item[28.] Two points each equidistant from the extremities of a line-segment
determine the perpendicular bisector of the segment.
\Unit{28.}%

\item[29.] One and only one perpendicular can be drawn to a line
through a point whether that point is on the line or not.
\Unit{29.}%

\item[30.] The sum of all consecutive angles about a point in a plane
is four right angles.
\Unit{30.}%

\item[31.] The sum of all consecutive angles about a point and on one
side of a straight line is two right angles.
\Unit{31.}%

\item[32.] If two adjacent angles are supplementary, their exterior
sides lie in the same straight line.
\Unit{32.}%

\item[33.] If in two triangles two sides of one are equal respectively
to two sides of the other, but the third side of the first is greater
than the third side of the second, then the included angle of the
first is greater than the included angle of the second.
\Unit{33.}%

\item[34.] Two lines which are perpendicular to the same line are
parallel.
\Unit{34.}%

\item[35.] If a line is perpendicular to one of two parallel lines, it is
perpendicular to the other also.
\Unit{35.}%

\item[36.] If two given lines are perpendicular respectively to each
of two intersecting lines, then the given lines are not parallel.

\item[37.] In a right triangle there are two acute angles.

\item[38.] From a point in a perpendicular to a straight line, oblique
segments are drawn to the line. Then,
\Unit{38.}%

\begin{Sublist}

\item[(1)] If the distances cut off from the foot of the perpendicular
are unequal, the oblique segments are unequal, that one being the
greater which cuts off the greater distance; and

\NewFolio %% -----File: 015.png---Folio 7-------

\item[(2)] Conversely, if the oblique segments are unequal, the distances
cut off are unequal, the greater segment cutting off the
greater distance.

\end{Sublist}

\item[39.] Two angles whose sides are perpendicular, each to each,
are equal or supplementary.

\item[40.] Two right triangles are equal if the hypotenuse and a side
of one are equal respectively to the hypotenuse and a side of the
other.
\Unit{40.}%

\item[41.] Two right triangles are equal if a side and an acute angle
of one are equal respectively to the corresponding side and acute
angle of the other.

\item[42.] Two right triangles are equal if the hypotenuse and an
acute angle of one are equal respectively to the hypotenuse and an
acute angle of the other.
\Unit{42.}%

\item[43.] A quadrilateral is a parallelogram
\Unit{43.}%

\begin{Sublist}

\item[(1)] if both pairs of opposite sides are equal; or

\item[(2)] if two opposite sides are equal and parallel.

\end{Sublist}

\item[44.] Opposite sides of a parallelogram are equal.

\item[45.] Two parallelograms are equal if an angle and the two
adjacent sides of one are equal respectively to an angle and the
two adjacent sides of the other.

\item[46.] The segment connecting the middle points of the two non-parallel
sides of a trapezoid is parallel to the bases and equal to
one half their sum.
\Unit{46.}%

\item[47.] The locus of all points equidistant from the extremities of
a line-segment is the perpendicular bisector of the segment.
\Unit{47.}%

\item[48.] In the same circle or in equal circles equal chords subtend
equal arcs.
\Unit{48.}%

\item[49.] A line perpendicular to a radius at its extremity is tangent
to the circle.

\item[50.] If a line is tangent to a circle, it is perpendicular to the
radius drawn to the point of contact.

\NewFolio %% -----File: 016.png---Folio 8-------

\item[51.] If in a proportion the antecedents are equal, then the consequents
are equal and conversely.

\item[52.] In a series of equal ratios the sum of any two or more
antecedents is to the sum of the corresponding consequents as any
antecedent is to its consequent.

\item[53.] If a line cuts two sides of a triangle and is parallel to the
third side, then any two pairs of corresponding segments form a
proportion.

\item[54.] If two sides of a triangle are cut by a line parallel to the
third side, a triangle is formed which is similar to the given
triangle.

\item[55.] In two similar triangles corresponding altitudes are proportional
to any two corresponding sides.

\item[56.] Two triangles are similar if an angle of one is equal to an
angle of the other and the pairs of adjacent sides are proportional.

\item[57.] Two triangles are similar if their pairs of corresponding
sides are proportional.

\item[58.] The area of a parallelogram is equal to the product of its
base and altitude.

\item[59.] Two parallelograms have equal areas if they have equal
bases and equal altitudes.

\item[60.] The area of a triangle is equal to one half the product of
its base and altitude.

\item[61.] If $a$ is a side of a triangle and~$h$ the altitude on it and~$b$
another side and~$k$ the altitude on it, then $ah = bk$.

\item[62.] The area of a trapezoid is equal to one half the product of
its altitude and the sum of its bases.
\Unit{62.}%

\item[63.] The area of a circle is one half the circumference times the
radius, or in symbols:
\[
a = \tfrac{1}{2} \cdot 2\pi r \cdot r = \pi r^2.
\]
\end{Theorems}
\index{Theorems@\textbf{Theorems of Plane Geometry}|)}%{\Pnum{63}}%

\NewFolio %% -----File: 017.png---Folio 9-------


\Chapter{BOOK I}{Properties of the Plane}

\Paragraph{64.}{Relations of Points, Lines, and Planes.} If a line or a plane
contains a point, the point is said to be \emph{on} the line or \emph{in} the
plane and the line or plane is said to pass \emph{through} the point.
If a plane contains a line, the line is said to be \emph{in} the plane
and the plane is said to pass \emph{through} the line.

\index{Axioms@\textbf{Axioms}|(}%{\Pnum{65}}%
\begin{Axiom}{65.}{1.}
If two points of a straight line lie in a plane
then the whole line lies in the plane.
\end{Axiom}

\begin{Remark}
Since a line is endless, it follows from this axiom that a plane is
endless in all its directions.
\end{Remark}

\begin{Axiom}{66.}{2.}
Through three non-collinear points one and
only one plane can be passed.
\end{Axiom}

\begin{Axiom}{67.}{3.}
Two distinct planes cannot \\
%[Illustration]
meet in one point only.\hfill\smash[t]{\raisebox{-0.125in}{\Input{017a}{1.25in}}}
\end{Axiom}
\index{Axioms@\textbf{Axioms}|)}%{\Pnum{67}}%


\Paragraph{68.}{Determination of a Plane.} A plane is said to be \emph{determined}
by certain elements (lines or points) if this plane contains
these elements while no other plane does contain them.
\index{Plane@\textbf{Plane}!determination of}%{\Pnum{68}}%

% [Illustration]
\begin{wrapfigure}[5]{r}{1.5in}
\vspace*{-0.375in}
\quad\Input{017b}{1.5in}
\end{wrapfigure}
\begin{Remark}
While two points determine a straight line it
is obvious that two points do not determine a plane.
The figure shows three planes, $L$, $M$, $N$, all passing
through the two points $A$~and $B$. But only a certain
one of these planes contains a given point~$C$
which is not in the line~$AB$.

We, therefore, say that three non-collinear points determine a plane,
while any number of collinear points fail to determine a plane.
\end{Remark}

\NewFolio %% -----File: 018.png---Folio 10-------

\Section{LINE COMMON TO TWO PLANES}
\SecToc{Perpendicular Planes and Lines}

\begin{Theorem}{69.}{I.}
Two intersecting planes meet in a
straight line.
\end{Theorem}
%[Illustration]
\begin{center}
\vspace*{-0.25in}
\Input{018a}{0.6\textwidth}
\end{center}
\vspace*{-0.25in}

\Objective{Given two intersecting planes $M$~and $N$.}
{To prove that they meet in a straight line~$AB$.}

\Proof: If two planes intersect they meet in at least two
points, as $A$~and $B$.\hfill Ax.~3, §~\Sref{67}

But $A$~and $B$ determine a line which lies wholly in $M$~and
also wholly in~$N$.\hfill Ax.~1, §~\Sref{65}

Hence the planes have the straight line~$AB$ in common.

A point~$C$ not in~$AB$ cannot lie in both $M$~and $N$, for in that
case the planes would have three non-collinear points in common
and hence would coincide.\hfill Ax.~2, §~\Sref{66}

Hence the planes $M$~and $N$ meet in the straight line~$AB$.
\QED % [** PP: Move \QED to same line]

\Paragraph{70.}{Foot of a Line Meeting a Plane.} The point in which a
\index{Foot@\textbf{Foot}!of a line}%{\Pnum{70}}%
straight line meets a plane is called the \emph{foot of the line}.

\Paragraph{71.}{Line and Plane Perpendicular to Each Other.} A line is
\index{Foot@\textbf{Foot}!of a perpendicular}%{\Pnum{71}}%
\index{Perpendicular@\textbf{Perpendicular}!line to a plane}%{\Pnum{71}}%
\index{Perpendicular@\textbf{Perpendicular}!plane to a line}%{\Pnum{71}}%
said to be \emph{perpendicular to a plane} if it is perpendicular to
every line in the plane passing through its foot. In this case
the plane is also said to be perpendicular to the line.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{72.}{Line and Plane Oblique to Each Other.} A line which meets
a plane and is not perpendicular to it is
said to be \emph{oblique} to the plane. The plane
is also said to be oblique to the line.
\end{minipage}%
%[Illustration]
\hfil\smash{\Input{018b}{1.25in}}\hfil


\begin{Remark}
\noindent In the figure, $PA$ is perpendicular to the plane~$M$
and $QA$ is oblique to it.
\end{Remark}

\NewFolio %% -----File: 019.png---Folio 11-------

\Section{ELEMENTS WHICH DETERMINE A PLANE}

\begin{Theorem}{73.}{II.}
A plane is determined by \emph{(1)} a
line and a point not on it, \emph{(2)} two intersecting lines,
and \emph{(3)} two parallel lines.
\end{Theorem}

% [Illustrations] [** PP: labels l_1 and l_2 reversed in third fig]
\Erratum{err019a}%
\begin{center}
\Input{019abc}{\textwidth}
\end{center}

\Objective{Given (1) a line~$l$ and a point~$P$ not on it; (2) two intersecting
lines $l_1$~and $l_2$; (3) two parallel lines $l_1$~and $l_2$.}
{To prove that in each case a plane is determined.}

% [Illustration]
\noindent\begin{minipage}[b]{\textwidth-1.375in}
\Indent\Proof: (1) Let $A$~and $B$ be two points on~$l$. Then
one and only one plane~$M$ can be passed through $l$~and
$P$ because one and only one plane can be passed
through $A$, $B$, and~$P$.\hfill \textsc{Ax}.~2, §~\Sref{66}

%[Illustration]

\Indent(2) Let $A$ be the intersection point of $l_1$~and $l_2$, and
$B$~and $C$ any other points, one on~$l_1$ and the other on~$l_2$.
Then $A$, $B$, and~$C$ determine the plane~$N$ in which lie
$l_1$~and $l_2$.\hfill \textsc{Axs}.~2, 1. §§~\Sref{66}, \Sref{65}

%[Illustration]

\Indent(3) By definition $l_1$~and $l_2$ lie in a plane~$R$. They
lie in only one such plane since the points $A$~and $B$
on~$l_1$ and~$C$ on~$l_2$ lie in only one plane.\footnoteTrans{See
\Tref{019a}{Transcriber's Notes}.}
% [** PP: labels l_1 and l_2 reversed from third fig]
\QED
\end{minipage}%
\hfill\smash{\Input{019def}{1.25in}}

\begin{Corollary}{74.}{1.}
Through a line there is more than one
plane.
\end{Corollary}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Suggestion}
Let $M$~be a plane through the given
line~$l$, and~$C$ a point not in~$M$. Then $l$~and $C$ determine
a plane~$N$ through~$l$ different from~$M$.
\end{Suggestion}

\Indent\begin{Corollary}{75.}{2.}
At a point on a line
there is more than one perpendicular to the line.
\end{Corollary}
\end{minipage}%
%[Illustration]
\hfill\smash{\Input{019g}{1.5in}}

\begin{Suggestion}
Let $M$~and $N$ be planes each passing through the given
line~$l$. Then in each plane there is a line~$\perp l$ at any point~$A$ on it.
\end{Suggestion}

\NewFolio %% -----File: 020.png---Folio 12-------

\Section{LINE PERPENDICULAR TO THE PLANE OF TWO GIVEN LINES}

\begin{Theorem}{76.}{III.}
If a line is perpendicular to
each of two lines at their point of intersection, it is
perpendicular to the plane of these lines.
\end{Theorem}

% [Illustration]
\begin{center}
\Input{020a}{0.6\textwidth}
\end{center}

\Objective{Given a line~$l$ perpendicular to each of the lines $l_1$~and $l_2$ at
the point~$P$.}
{To prove that~$l$ is perpendicular to the plane of $l_1$~and $l_2$.}

\Proof: Let $M$ be the plane of $l_1$~and $l_2$, and let~$l_3$ be any line
in~$M$ through~$P$. Draw a line meeting $l_1$, $l_2$, and~$l_3$ in the points
$B$, $C$, and~$D$ respectively. Let $E$~and $F$ be points on~$l$, on
opposite sides of~$P$, and such that $EP=FP$. Draw $EB$, $ED$,
$EC$, $FB$, $FD$, $FC$.

Then prove:
\begin{align*}
\text{(1)} \bigtriangleup EBP &= \bigtriangleup FBP; \quad
\text{(2)} \bigtriangleup ECP = \bigtriangleup FCP; \\
\text{(3)} \bigtriangleup EBC &= \bigtriangleup FBC; \quad
\text{(4)} \bigtriangleup EBD = \bigtriangleup FBD; \\
\text{(5)} \bigtriangleup EPD &= \bigtriangleup FPD; \quad
\text{(6)} \,\angle\, EPD = \angle\, FPD. \\
\therefore\ & EP \text{ is perpendicular to~$l_3$.} \tag*{Why?}
\end{align*}

But $l_3$ is any line in~$M$ through~$P$. \\
\noindent\null\hfill$\therefore$\ line $l \perp$ plane $M$.\hfill §~\Sref{71} \\
\QED

\begin{Corollary}{77.}{}
If each of two lines is perpendicular to a
third line at the same point, then the plane of the two lines is
perpendicular to the third line.
\end{Corollary}

\NewFolio %% -----File: 021.png---Folio 13-------

\begin{Sightwork}

\item[] The diagram on this page represents a three-dimensional figure in the
shape of an ordinary rectangular box. In this figure the points $A$, $K$,
and~$B$, for instance, do not determine a plane, since they all lie on the
same straight line, while $A$, $B$, and~$C$ do not lie in a straight line, and
hence determine a plane.

\item[1.] In this figure pick out several lines which lie in one of the surfaces
and are not obscured by the figure.

\item[2.] Pick out several lines which are obscured by the figure; also some
which lie within the figure.

\item[3.] Pick out four sets of three points each which do not determine
planes, and also four sets which do determine planes.

% [Illustration]
\begin{center}
\Input{021a}{0.5\textwidth}
\end{center}

\item[4.] Is the line~$AB$ perpendicular to the plane~$BCG$? Why? Is $AB$
perpendicular to the plane $AEH$?

\item[5.] Pick out six planes in the figure, each determined by parallel lines.

\item[6.] Do the points $C$, $Z$, $E$ determine a plane? the points $C$, $Z$, $G$?
the points $B$, $F$, $Z$?

\item[7.] Using the schoolroom, or a room at home, locate planes corresponding
to the planes $AEG$, $KLM$, $NOP$, and~$EFG$, in the above
figure.

\item[8.] Point out in some room planes determined by points corresponding
to $D$, $E$, $B$; $D$, $F$, $B$; $D$, $C$, $F$; $A$, $B$, $H$ in the above figure.

\end{Sightwork}

\NewFolio %% -----File: 022.png---Folio 14-------


\Section{PLANE PERPENDICULAR TO A LINE}

\begin{Theorem}{78.}{IV.}
Through a point there is one
and only one plane perpendicular to a line.
\end{Theorem}

% [Illustration: FIG. 1.]      [Illustration: FIG. 2.]
\vspace*{-0.125in}
\begin{center}
\Input{022ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given a line~$l$ and a point~$P$.}
{To prove that through~$P$ there is one and only one plane $\perp l$.}

\Proof: \Case{(1)}{When the point~$P$ is on the line~$l$.}\quad \textsc{Fig}.~1.

Through $P$ draw lines $PQ$~and $PQ'$ both $\perp l$. Then the
plane~$M$, determined by $PQ$~and $PQ'$, is $\perp l$.\hfill §~\Sref{77}

To prove that $M$ is the only plane through~$P$ which is $\perp l$,
suppose that a plane~$M'$ through~$P$ is also $\perp l$. Let $R$ be a
plane through~$l$ meeting $M$~and $M'$ in two lines. Then these
lines would both lie in~$R$ and be $\perp l$, which is impossible.\hfill §~\Sref{29}

\Case{(2)}{When the point~$P$ is not on the line~$l$.}\quad \textsc{Fig}.~2.

Let $PQ$ be a line $\perp l$ and let $N$ be a plane $\perp l$ at~$Q$.

Then the plane~$N$ passes through~$P$ and is the plane
required.

For if $P$ does not lie in~$N$, then a plane~$R'$ determined by $l$~and
$PQ$ cuts~$N$ in a line~$l'$ which is $\perp l$ (§~\Sref{71}), and $PQ$~and $l'$ are
each $\perp l$ at the point~$Q$ in plane~$R'$, which is impossible.\hfill §~\Sref{29}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent Suppose, now, that there are two planes
through~$P$ each $\perp l$. These planes cannot
meet~$l$ in the same point (Case 1). Let them
meet~$l$ in $Q$~and $Q'$. Then $PQ$~and $PQ'$ are
each $\perp l$, which is impossible.
\QED
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{022c}{1.5in}}

\begin{Corollary}{79.}{}
All lines perpendicular to a line at the same
point lie in the plane perpendicular to the line at this point.
\end{Corollary}

\NewFolio %% -----File: 023.png---Folio 15-------

\Section{LINE PERPENDICULAR TO A PLANE}

\begin{Theorem}{80.}{V.}
Through a point there is one and
only one line perpendicular to a plane.
\end{Theorem}

% [Illustrations]
\begin{center}
\vspace*{-0.25in}
\Input{023ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given a plane~$M$ and a point~$P$.}
{To prove that through~$P$ there is one and only one line $\perp M$.}

\Proof: \Case{(1)}{When $P$ is in the plane~$M$} (first figure). Let $l_1$~be any line
in~$M$ through~$P$, and let $N$ be a plane $\perp l_1$ at~$P$ and meeting~$M$ in the
line~$l_2$. Let $PA$ be a line in~$N$ and $\perp l_2$. Then $PA$ is also $\perp l_1$.\hfill §~\Sref{71}
\[
\therefore PA \perp M. \tag*{§~\Sref{76}.}
\]

\Case{(2)}{When $P$ is not in the plane~$M$} (second figure). Let~$l$ be any line
in the plane~$M$. Through~$P$ pass a plane~$N' \perp l$ at~$A$ and meeting~$M$ in
the line~$AK$. From~$P$ in plane~$N'$ draw a line~$PO \perp AK$ and extend it
to~$P'$ so that $OP' = PO$. Let $B$ be any point in~$l$ different from~$A$.
Draw $PB$, $PA$, $P'B$, $P'A$, and~$OB$.

Then prove (1)~$\triangle POA = \triangle P'OA$;\quad (2)~$\triangle PAB = \triangle P'AB$;\\
(3)~$\triangle POB = \triangle P'OB$;\quad (4)~$\angle POB = \angle P'OB$;\quad (5)~$PO \perp OB$.
\[
\therefore PO \perp M. \tag*{§~\Sref{76}}
\]

% [Illustrations]
\begin{wrapfigure}[7]{r}{3in}
\vspace*{-0.25in}
\Input{023cd}{3in}
\end{wrapfigure}
If in either case~(1) or case~(2) there were two lines $PA$~and
$PB$ each $\perp M$, then the
plane~$R$ of these lines would
cut~$M$ in a line~$l$. Hence
$PA$~and $PB$ would both
lie in~$R$ and be $\perp l$, which
also lies in~$R$. But this is
impossible by §~\Sref{29}.

Hence $PA$ is the only line through~$P$ which is $\perp M$.
\QED

\NewFolio %% -----File: 024.png---Folio 16-------

\begin{Sightwork}

\item[1.] Does a stool with three legs always stand firmly on a flat floor?
Why?

\item[2.] Does a table with four legs always stand firmly on a flat floor?
Why? On what conditions will such a table stand
firmly on a flat floor?

\begin{minipage}[b]{3in}
\item[3.] If the point $C$ does not lie in the plane $ABD$,
how many different planes are determined by the
points $A$, $B$, $C$, $D$?

\item[4.] How many planes are determined by any four
points which do not all lie in one plane?
\end{minipage}%%
% [** Illustration]
\hfil\Input{024a}{1.5in}\hfil

\item[5.] How many planes are determined by the points $A$, $B$, $C$, $D$, $E$, if
$A$, $B$, $C$ lie in a straight line and $C$, $D$, $E$ lie in another straight line?

\item[6.] How many planes are determined by five points, no four of which
lie in the same plane?

\begin{minipage}[b]{\textwidth-1.25in}
\item[7.] How many planes are determined by three lines
$l_1$, $l_2$, $l_3$ all passing through the same point but not all
lying in the same plane?

\item[8.] How many planes are determined by four lines
which all meet in a point, but no three of which lie in
the same plane?

\item[9.] How many planes are determined by three lines
all parallel to each other, and not all lying in the same
plane?
\end{minipage}%%
% [** Illustration]
\hfil\Input{024b}{1in}\hfil


\item[10.] How many planes are determined by four lines
all parallel to each other, and no three lying in the same plane?

\item[11.] A line cannot be perpendicular to each of two intersecting planes.
Why?
\end{Sightwork}

\noindent\begin{minipage}[b]{\textwidth-2.5in}
\Indent\begin{Suggestion}[s]
(1)~If a line $l$ is perpendicular to the planes $M$ and $N$ at
the points $A$ and $B$, and $C$ is a point
in their intersection, then $\triangle ABC$
would contain two right angles.
\end{Suggestion}
\end{minipage}%
\hfill\Input{024cd}{2.25in}

{\small (2)~If $l$ is perpendicular to $M$ and $N$
at a point $P$ in their intersection,
pass a plane through $l$, meeting $M$
and $N$ in $l_1$ and $l_2$. Then in this plane $l_1$ and $l_2$ are both $\perp l$.}

\NewFolio %% -----File: 025.png---Folio 17-------

\Section{OBLIQUE LINES FROM A POINT TO A PLANE}

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\begin{Theorem}{81.}{VI.}
Oblique lines from a point to a
plane meeting the plane at equal distances from the
foot of the perpendicular are equal; and

\Indent conversely, two equal oblique lines from a point to
a plane meet the plane at equal distances from the foot
of the perpendicular.
\end{Theorem}
\end{minipage}%
\hfill
% [Illustration]
\raisebox{0.25in}{\Input{025a}{2in}}

\Objective{(1) Given $PC \perp M$, and $AC = BC$.}{To prove that $PA = PB$.}

\begin{Suggestion}
Suggestion. Prove $\triangle PCA = \triangle PCB$.
\end{Suggestion}

\Objective{(2) Given $PC \perp M$, and $PA = PB$.}{To prove that $AC = BC$.}

\begin{Suggestion}
Suggestion. Prove $\triangle PCA = \triangle PCB$.
\end{Suggestion}

\begin{Corollary}{82.}{}
The perpendicular is the shortest distance
from a point to a plane.
\end{Corollary}

Hence \emph{the distance} from a point to a plane means the perpendicular
distance.

\begin{Sightwork}

\item[] Without giving proofs describe the following loci:

\item[1.] All points equidistant from the points on a circle.

\item[2.] All points equidistant from the vertices of a triangle.

\item[3.] All points in a plane which are at a given distance from a given
point outside the plane. If a perpendicular be drawn to the plane from
this outside point, how is its foot related to this locus?
\end{Sightwork}

\NewFolio %% -----File: 026.png---Folio 18-------

\begin{Exercises}

\item[1.] Show how a carpenter could use the theorem of §~\Sref{76} to
stand a post perpendicular to the floor, if he has at hand two
ordinary steel squares.

\noindent\begin{minipage}[b]{\textwidth-2.5in}
\Indent\item[2.] Show how a back-stop on a ball field can be made perpendicular
to the line through
second base and the home plate.
What theorems of solid geometry
are used?
\end{minipage}%%
% [** Illustration]
\hfil\smash[t]{\Input{026a}{2.25in}}\hfil

\item[3.] If a plane is perpendicular
to a line-segment~$PP'$ at its
middle point, prove: (1) Every
point in the plane is equally distant from $P$ and~$P'$; (2) every
point equally distant from $P$ and~$P'$ lies in this plane. What
is the locus of all points in space equidistant from $P$ and~$P'$?
Compare §~\Sref{47}.

\item[4.] Given the points $A$ and~$B$ not in a plane~$M$. Find the
locus of all points in $M$ equidistant from $A$ and~$B$.

\begin{Suggestion}
All such points must lie in the plane $M$ and also in the
plane which is the perpendicular bisector of the segment $AB$.
\end{Suggestion}

\item[5.] Find the locus of all points equidistant from two given
points $A$ and $B$, and also equidistant from two points $C$ and $D$.
Discuss.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\item[6.] State and prove a theorem of solid
geometry corresponding to the theorem of
plane geometry given in §~\Sref{38}.
% [** Illustration]

\item[7.] If in the figure $PD \perp$ plane $M$, and
$DC \perp AB$, a line of the plane $M$, prove that
$PC \perp AB$.

% [** Illustration]

\Indent\begin{Suggestion}
Lay off $CA = CB$, and compare triangles.
\end{Suggestion}

\item[8.] If in the same figure $PD \perp M$, and
$PC \perp AB$, a line of the plane, prove that $DC \perp AB$.
\end{minipage}%
\hfil\smash{\raisebox{-0.25in}{\Input{026bc}{1.25in}}}\hfil

\end{Exercises}

\NewFolio %% -----File: 027.png---Folio 19-------

\Section{PARALLEL LINES PERPENDICULAR TO THE SAME PLANE}

\begin{Theorem}{83.}{VII.} Two lines perpendicular to the
same plane are parallel; and

Conversely, if one of two parallel lines is perpendicular
to a plane, the other is also.
\end{Theorem}

% [** Illustration Fig. 1.]
% [** Illustration Fig. 2.]
\begin{center}
\Input{027ab}{}
\end{center}

\Objective%
{Given (1) $AB$ and $CD$ each $\perp$ the plane $M$.\quad \textsc{Fig}.~1.}
{To prove that $AB \parallel CD$.}

\Proof: Draw $BD$ and make $DE \perp DB$.

Take points $A$ and $E$ so that $BA = DE$, and draw $AD$, $AE$,
and $BE$.

Now prove: (1)~$\triangle ABD = \triangle BDE$ and $\therefore AD = BE$;

(2)~$\triangle ADE = \triangle ABE$ and $\therefore\angle ADE = \angle ABE = \text{Rt.}~\angle$.

$\therefore DC$, $DA$, $DB$, and $BA$ all lie in the same plane.\hfill §~\Sref{79} \\
\null\hfill $AB \parallel CD$.\hfill §~\Sref{34}

\Objective{Given (2) $AB \parallel CD$ and $AB \perp M$.\quad \textsc{Fig}.~2.}
          {To prove that $CD \perp M$.}

\Proof: If $CD$ is not $\perp M$ let $C'D$ be $\perp M$.

Then $C'D \parallel AB$ by case (1), and $C'D$ coincides with $CD$ §~\Sref{21}
and $\therefore CD \perp M$.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Corollary}{84.}{1.}
If each of two lines is
parallel to a third line they are parallel to each
other.
\end{Corollary}

\Indent\begin{Corollary}{85.}{2.}
If a plane is perpendicular
to one of two parallel lines, it is perpendicular
to the other.
\end{Corollary}
\end{minipage}%%
% [** Illustration]
\hfill\raisebox{0.25in}{\Input{027c}{1.25in}}


\NewFolio %% -----File: 028.png---Folio 20-------

\Section{PARALLEL PLANES AND LINES}
\SecToc{Parallel Planes and Lines}

\Paragraph{86.}{Parallel Planes.} Two planes which do not meet are said
to be \emph{parallel}.
\index{Parallel@\textbf{Parallel}!plane to a plane}%{\Pnum{86}}%

\Paragraph{87.}{Line Parallel to a Plane.} A straight line and a plane
which do not meet are said to be parallel.
\index{Parallel@\textbf{Parallel}!line to a plane}%{\Pnum{87}}%

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{88.}{Intercepted Segments.} If a
straight line $l_2$ meets two planes in $A$
and $B$, then the segment $AB$ is said
to be \emph{intercepted} by the planes.
\index{Segment@\textbf{Segment}!intercepted by planes}%{\Pnum{88}}%

\Indent\begin{Remark}
Any line, as $l_1$, in either of two parallel
planes, $M$ and $N$, is parallel to the other
plane. The segment $AB$ on the line $l_2$
is intercepted by the planes.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfill\smash[b]{\raisebox{-0.25in}{\Input{028a}{2in}}}


\Section{LINE PARALLEL TO A PLANE}

\begin{Theorem}{89.}{VIII.} If a straight line is parallel
to a given plane, it is parallel to the intersection of
any plane through it with the given plane.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{028b}{0.6\textwidth}
\end{center}

\begin{Suggestion}[ for proof]
If $l_1$ is the given line, $M$ the given plane, and
$l_2$ the intersection of a plane $N$ through $l_1$ with $M$, show that $l_1$ and $l_2$ lie
in plane $N$ and cannot meet.
\end{Suggestion}

\begin{Corollary}{90.}{1.}
If a line outside a plane is parallel to
some line in the plane, then the first line is parallel to the plane.
\end{Corollary}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Corollary}{91.}{2.}
If a line is parallel to
a plane, then through any point in the plane
there is a line in the plane parallel to the given
line.
\end{Corollary}

\Indent\begin{Corollary}{92.}{3.}
The intersections of a plane
with two parallel planes are parallel lines.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{028c}{1.5in}}


\NewFolio %% -----File: 029.png---Folio 21-------

\Section{PLANES PERPENDICULAR TO A LINE ARE PARALLEL}

\begin{Theorem}{93.}{IX.}
If each of two planes is perpendicular
to the same line, they are parallel; and

Conversely, if one of two parallel planes is perpendicular
to a line, the other is also.
\end{Theorem}
% [** Illustration: FIG. 1.]
% [** Illustration: FIG. 2.]
\begin{center}
\Input{029ab}{}
\end{center}

\Objective%
{Given (1) plane $M\perp AB$ and plane $N\perp AB$. \textsc{Fig}.~1.}
{To prove that $M\parallel N$.}

\Proof: Suppose $M$ and $N$ to meet in some point $P$. Draw
$AP$ in $M$ and $BP$ in $N$. Then $AB\perp AP$ and $AB\perp BP$ (§~\Sref{71}),
which is impossible.\hfill §~\Sref{29}

\Objective%
{Given (2) $M\parallel N$ and $M\parallel AB$. \textsc{Fig}.~2.}
{To prove that $N\parallel AB$.}

\Proof: Through $AB$ pass a plane cutting $M$ and $N$ in $AC$
and $BD$, and a second plane cutting $M$ and $N$ in $AE$ and $BF$.

Then, $AC\parallel BD$ and $AB\parallel BF$.\hfill §~\Sref{92}

Now prove (1) $AB\perp BD$, (2) $AB\perp BF$.\hfill §~\Sref{35}

\null\hfill $AB\perp N$.\hfill §~\Sref{76}

\begin{Corollary}{94.}{1.}
Parallel line-segments included between
parallel planes are equal.
\end{Corollary}

\begin{Corollary}{95.}{2.}
If a line is perpendicular to one of two
parallel planes, it is perpendicular to the other also.
\end{Corollary}

\begin{Corollary}{96.}{3.}
Two planes each parallel to a third plane
are parallel to each other.
\end{Corollary}

\NewFolio %% -----File: 030.png---Folio 22-------

\Section{PARALLEL PLANES}

\begin{Theorem}{97.}{X.}
If a plane is parallel to each of
two intersecting lines, it is parallel to the plane of these
lines.
\end{Theorem}
% [** Illustration]
\begin{center}
\Input{030a}{0.6\textwidth}
\end{center}

\Objective%
{Given a plane $M$ parallel to the intersecting lines $l_1$ and $l_2$.}
{To prove that the plane $M$ is $\parallel$ the plane $N$ of $l_1$ and $l_2$.}

\Proof: If $M$ is not $\parallel N$, these planes meet in a line $l_3$.
\hfill §~\Sref{69}

Then neither $l_1$ nor $l_2$ can meet $l_3$, since they are $\parallel M$.

Hence $l_1\parallel l_3$, and $l_2\parallel l_3$, which is impossible.
\hfill §~\Sref{21}

$\therefore M$ and $N$ cannot meet and are parallel.
\QED

\begin{Theorem}{98.}{XI.}
Through a point not in a plane
there is one and only one plane parallel to this plane.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Objective{\Indent Given a plane $M$ and a point~$P$ not in~$M$.}
          {\Indent To prove that through~$P$ there is one and only
           one plane $\parallel M$.}
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.125in}{\Input{030b}{1.75in}}}

\Proof: Let $l$ be a line through $P$ and $\perp M$.

Through $P$ draw $l_1$ and $l_2$ each $\perp l$.\hfill §~\Sref{75}

The plane $N$ of $l_1$ and $l_2$ is $\perp l$, and hence $N\parallel M$.
\hfill §§~\Sref{76}, \Sref{93}

If through $P$ there were another plane $R\parallel M$, then
$R$ would be $\perp l$ at~$P$.
\hfill §~\Sref{93}

But $N$ and $R$ cannot both be $\perp l$ at $P$.
\hfill §~\Sref{78}

Hence $N$ is the only plane through $P\parallel M$.
\QED % [** PP: Move \QED to same line]

\NewFolio %% -----File: 031.png---Folio 23-------

\Section{PLANES PARALLEL TO GIVEN LINES}

\begin{Theorem}{99.}{XII.}
Through one of two skew lines
there is one and only one plane parallel to the other line.
\end{Theorem}
% [** Illustration]
\begin{center}
\Input{031a}{0.6\textwidth}
\end{center}

\Objective%
{Given two skew lines $l_1$ and $l_2$. See §~\Sref{7}.}
{To prove that through $l_1$ there is one and only one plane $\parallel l_2$.}

\Proof: Through $P$, a point in $l_1$, draw a line $l_3\parallel l_2$.

Then $l_1$ and $l_3$ determine a plane $M\parallel l_2$.\hfill  §~\Sref{90}

Any other plane $N$ through $l_1$ would meet the plane of $l_3$ and
$l_2$ in a line through $P$ not $\parallel l_2$ and hence $N$ would meet $l_2$. \QED % [** PP: Move \QED to same line]

\begin{Theorem}{100.}{XIII.}
Through a point outside of
each of two non-parallel lines there is one and only one
plane parallel to both of these lines.
\end{Theorem}
% [** Illustration]
\begin{center}
\Input{031b}{0.6\textwidth}
\end{center}

\Objective%
{Given a point $P$ outside of the non-parallel lines $l_1$ and $l_2$.}
{To prove that there is one and only one plane $M$ through $P$
parallel to $l_1$ and~$l_2$.}

\Proof: Through $P$ pass $l_3\parallel l_1$ and $l_4\parallel l_2$.

Then the plane $M$ of $l_3$ and $l_4$ is parallel to $l_1$ and $l_2$.
\hfill §~\Sref{90}

In any other plane $N$ through $P\parallel l_1$ and $l_2$, there are lines
$l'_3$ and $l'_4$ through $P$ such that $l'_3\parallel l_1$ and $l'_4\parallel l_2$.
\hfill §~\Sref{91}

In that case $l'_3$ is identical with $l_3$ and $l'_4$ with $l_4$.
\hfill §~\Sref{21}

Hence any such plane $N$ is identical with $M$.
\QED

\NewFolio %% -----File: 032.png---Folio 24-------

\begin{Exercises}

\item[1.] Given a plane $M$ and a point $P$ not in $M$. Find the locus
of the middle points of all segments connecting $P$ with points
in $M$.

\begin{Suggestion}
Use the fact that a line parallel to the base of a triangle
and bisecting one side bisects the other side also.
\end{Suggestion}

\item[2.] Show that a plane containing one only of two parallel
lines is parallel to the other.

\item[3.] If in two intersecting planes a line of one is parallel to
a line of the other, then each of these lines is parallel to the
line of intersection of the planes.

\item[4.] Show that three lines which do not meet in one point
must all lie in the same plane if each intersects the other two.

\item[5.] Show that three planes, each of which intersects the
other two, have a point in common unless their three lines of
intersection are parallel.

\begin{Suggestion}
Suppose two of the intersection lines are not parallel, but
meet in some point $O$. Then show that the other line of intersection
passes through $O$, and hence that $O$ is the point common to all three
planes.
\end{Suggestion}

\item[6.] Given two intersecting planes $M$ and $N$. Find the locus
of all points in $M$ at a given perpendicular distance from $N$.

\item[7.] Given two non-intersecting lines $l_1$ and $l_2$. Find the locus
of all lines meeting $l_1$ and parallel to $l_2$.

\item[8.] Prove that the middle points of the sides of any quadrilateral
in space are the vertices of a parallelogram.

\begin{Suggestion}
Use the fact that a line bisecting two sides of a triangle
is parallel to the third side. Note that the four vertices of a quadrilateral
in space do not necessarily all lie in the same plane.
\end{Suggestion}

\item[] State the corresponding theorem in plane geometry.

\item[9.] In erecting a flagpole on a level space, show how it can
be made perpendicular by means of three ropes of equal length.
See §~\Sref{81}.
\end{Exercises}

\NewFolio %% -----File: 033.png---Folio 25-------

\Section{ANGLES WHOSE SIDES ARE PARALLEL}

\begin{Theorem}{101.}{XIV.} If two intersecting lines in
one plane are parallel, respectively, to two intersecting
lines in another plane, then the two planes are parallel,
and the corresponding angles formed by the lines are
equal.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{033a}{0.6\textwidth}
\end{center}

\Objective%
{Given the planes $M$ and $N$ in which $AB\parallel A'B'$, and $AC\parallel A'C'$.}
{To prove that $M\parallel N$ and $\angle 1 = \angle 2$.}

\Proof: (1) If $M$ is not $\parallel N$, then these planes meet in a
line $l$.
\hfill Why?

Then neither $AB$ nor $AC$ can meet $l$ since each is $\parallel N$.

$\therefore AB\parallel l$ and $AC\parallel l$ which is impossible.
\hfill Why?

(2) To prove $\angle 1 = \angle 2$, lay off $AB = A'B'$, $AC = A'C'$, and
draw $BC$, $B'C'$, $AA'$, $BB'$, and $CC'$.

\textit{Analysis}: $\angle 1 = \angle 2$ if $\triangle ABC = \triangle A'B'C'$, which is true
if $BC = B'C'$.

But $BC=B'C'$ if $BB'C'C$ is a $\Pgram$, which is so if $BB' = CC'$
and $BB' \parallel CC'$. This last is true if $AA'C'C$ and $AA'B'B$ are
$\Pgrams$, for then $BB' = AA' = CC'$ and $BB'\parallel AA'\parallel CC'$.

Hence we need to prove in order (1) $AA'C'C$ and $AA'B'B$
are $\Pgrams$, (2) $BB'C'C$ is a $\Pgram$, (3) $\triangle ABC=\triangle A'B'C'$, and
(4) $\angle BAC = \angle B'A'C'$.

Hence it is proved that the planes are parallel and the angles
are equal.
\QED

\begin{Corollary}{102.}{}
Two angles in space whose sides are
parallel each to each are either equal or supplementary.
\end{Corollary}

\NewFolio %% -----File: 034.png---Folio 26-------

\Section{PARALLEL PLANES INTERCEPT PROPORTIONAL SEGMENTS}

\begin{Theorem}{103.}{XV.}
If two straight lines are cut
by three parallel planes, the intercepted segments on
one line are proportional to the corresponding segments
on the other.
\end{Theorem}
% [** Illustration]
\begin{center}
\Input{034a}{0.4\textwidth}
\end{center}

\Objective%
{Given the lines $AB$ and $CD$ cut by the planes $M$, $N$, and~$P$.}
{To prove that $\dfrac{AE}{EB} = \dfrac{CG}{GD}$.}

\textbf{Outline of Proof}: Draw $AD$ and let the plane determined by
$AD$ and $CD$ cut the planes $M$ and $N$ in $AC$ and $FG$ respectively;
and let the plane of $AB$ and $AD$ cut $N$ and $P$ in $EF$
and $BD$ respectively.

Then prove (1) $FG\parallel AC$ and $EF\parallel BD$,
\[
(2)\quad \dfrac{AE}{EB} = \dfrac{AF}{FD} \text{ and }
         \dfrac{CG}{GD} = \dfrac{AF}{FD},\qquad
(3)\quad \dfrac{AE}{EB} = \dfrac{CG}{GD}.
\]

\begin{Corollary}{104.}{} Parallel planes which intercept equal segments
on any transversal line intercept equal segments on every
transversal line.
\end{Corollary}

\begin{Sightwork}

\item[] Review the theorems of solid geometry up to this point by stating the
corresponding theorems of plane geometry, in case such theorems exist.
\end{Sightwork}

\NewFolio %% -----File: 035.png---Folio 27-------

\begin{Exercises}

\item[1.] If the spaces between four parallel shelves are $5$, $8$, and
$10$ inches respectively, and a slanting rod intersecting them
has a $7$-inch segment between the first two shelves, find the
other two segments of the rod.

\item[2.] If a line cuts three parallel planes, $M$, $N$, $R$, so that the
segment intercepted between $M$ and $N$ is $7$ and that between
$N$ and $R$ is $21$, and if another line cuts the same planes so
that the segment between $M$ and $N$ is $11$, find the segment
on the second line between $N$ and $R$.

\item[3.] Show that line-segments included between parallel
planes and perpendicular to them are equal, and hence that
parallel planes are everywhere equally distant. How can a
carpenter make use of this principle in placing two parallel
shelves? How many distances must he measure? Why?

\item[4.] Show that through a point outside a plane any number
of lines can be drawn parallel to the plane. How are all
these parallels related?

\item[5.] Prove that if a plane bisects two sides of a triangle it is
parallel to the third side.

\item[6.] The perpendicular distance from a point $P$ to a plane
is $12$~in. Find the radius of the circle which is the locus of
all points in the plane at a distance of $20$~in.\ from $P$.

\item[7.] Show that, if three line-segments not in the same plane
are equal and parallel, the triangles formed by joining their
extremities, as in the figure of §~\Sref{101}, are equal and their
planes are parallel.

\item[8.] What is the relation of two lines if they are (a)~parallel
to a given line, (b)~perpendicular to a given line, (c)~parallel to
a given plane, (d)~perpendicular to a given plane?

\item[9.] What is the relation of two planes if they are both
(a)~parallel to a given plane, (b)~parallel to a given line,
(c)~perpendicular to a given line?
\end{Exercises}

\NewFolio %% -----File: 036.png---Folio 28-------

\Section{DIHEDRAL ANGLES}
\SecToc{Dihedral Angles}

\Paragraph{105.}{Dihedral Angle.} The part of a plane on one side of a
\index{Angle@\textbf{Angle}!dihedral}%{\Pnum{105}}%
\index{Dihedral@\textbf{Dihedral angles}}%{\Pnum{105}}%
\index{Edge@\textbf{Edge}!of a dihedral angle}%{\Pnum{105}}%
\index{Faces@\textbf{Faces}!of a dihedral angle}%{\Pnum{105}}%
\index{Halfplane@\textbf{Half-plane}}%{\Pnum{105}}%
line in it is called a \emph{half-plane}. The line is called the \emph{edge} of
the half-plane. Two half-planes meeting in a common edge
form a \emph{dihedral angle}. The common edge is the \emph{edge} of the
angle and the half-planes are its \emph{faces}.

\Paragraph{106.}{Plane Angle of a Dihedral Angle.} Two lines in the respective
\index{Dihedral@\textbf{Dihedral angles}!plane angle of of}%{\Pnum{106}}%
\index{Plane Angle@\textbf{Plane Angle}!of a dihedral angle}%{\Pnum{106}}%
faces of a dihedral angle and perpendicular to its
edge at a common point form a plane angle, which
is called \emph{the plane angle of the dihedral angle}.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\begin{Remark}
In the figure, the half-planes $M$ and $N$ form the dihedral
angle $M$-$AB$-$N$, read by naming the two faces and the
edge. If $CD$ in $N$ is $\perp AB$ and $ED$ in $M$ is $\perp AB$, then
$\angle CDE$ is the plane angle of the dihedral angle $M$-$AB$-$N$.
\end{Remark}

% [** Illustration]

By §~\Sref{101} all plane angles of a dihedral angle
are equal to each other.

\Indent\Paragraph{107.}{Generation of a Dihedral Angle.} A dihedral
\index{Dihedral@\textbf{Dihedral angles}!generation of}%{\Pnum{107}}%
angle may be thought of as \emph{generated} by the
rotation of a half-plane about its edge. The
magnitude of the angle depends solely upon the
\emph{amount of rotation}.
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.125in}{\Input{036ab}{1in}}}

\Paragraph{108.}{Equal Dihedral Angles.} Two dihedral angles are \emph{equal}
\index{Dihedral@\textbf{Dihedral angles}!equal}%{\Pnum{108}}%
when they can be so placed that they coincide.

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{109.}{Right Dihedral Angle.} A right
\index{Dihedral@\textbf{Dihedral angles}!right}%{\Pnum{109}}%
dihedral angle is one whose \emph{plane angle}
is a right angle.

% [** Illustration]
\Indent\Paragraph{110.}{Perpendicular Planes.} Two
planes are said to be \emph{mutually perpendicular}
if their dihedral angle is a right angle.
\index{Perpendicular@\textbf{Perpendicular}!plane to a plane}%{\Pnum{110}}%
\end{minipage}%
\hfill\smash[t]{\raisebox{0.25in}{\Input{036c}{2in}}}

\begin{Remark}
Dihedral angles are \emph{acute} or \emph{obtuse} according as their plane angles
are acute or obtuse.

Two dihedral angles are \emph{adjacent}, \emph{vertical}, \emph{supplementary}, or \emph{complementary}
according as their plane angles possess these properties.
\end{Remark}

\NewFolio %% -----File: 037.png---Folio 29-------

\Section{DIHEDRAL ANGLES AND THEIR PLANE ANGLES}

\begin{Theorem}{111.}{XVI.}
Two dihedral angles are
equal if their plane angles are equal; and

Conversely, if two dihedral angles are equal, their
plane angles are equal.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{037ab}{0.6\textwidth}
\end{center}

\Objective{Given (1) the dihedral angles $M-AB-N$ and $M'-A'B'-N'$ in
which the plane $\Angles CDE$ and $C'D'E'$ are equal.}
{To prove that $\angle M-AB-N = \angle M'-A'B'-N'$.}

\Proof: Place the equal $\Angles CDE$ and $C'D'E'$ in coincidence.

Then $AB$ coincides with $A'B'$ since these lines are perpendicular
to the plane $CDE$ at the point~$D$.\hfill              §~\Sref{80}

Then $M$ and $M'$ coincide as do also $N$ and $N'$ since they are
determined by coincident lines.\hfill                 §~\Sref{73}
\[
\therefore\ \angle M-AB-N = \angle M'-A'B'-N'. \tag*{§~\Sref{108}}
\]

\Objective{Given (2) $\angle M-AB-N = \angle M'-A'B'-N'$.}
{To prove that the plane $\Angles CDE$ and $C'D'E'$ are equal.}

\Proof: Place the equal dihedral $\Angles M-AB-N$ and $M'-A'B'-N$
in coincidence so that the points $D$ and $D'$ coincide.

Then $CD$ and $C'D'$ coincide as do also $ED$ and $E'D'$.
\hfill §~\Sref{29}
\[
\therefore\ \angle CDE = \angle C'D'E'. \tag*{\QED}
\]

\begin{Corollary}{112.}{}
All right dihedral angles are equal.
\end{Corollary}

\Paragraph{113.}{Measure of a Dihedral Angle.} It follows from §~\Sref{111}
\index{Dihedral@\textbf{Dihedral angles}!measure of}%{\Pnum{113}}%
that the plane angle of a dihedral angle may be regarded as
its measure.

\NewFolio %% -----File: 038.png---Folio 30-------

\Section{MUTUALLY PERPENDICULAR PLANES}

\begin{Theorem}{114.}{XVII.}
If two planes are mutually
perpendicular, and if a line is drawn from a point in
one perpendicular to their intersection, then this line
is perpendicular to the second plane.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{038a}{0.4\textwidth}
\end{center}

\Objective%
{Given plane $M \perp$ plane $N$ and a point $P$ in~$M$. Let $l_1$ be the
intersection of $M$ and $N$ and $l$ a line in $M$ through $P\perp l_1$.}
{To prove that $l\perp N$.}

\Proof: Let $l$ meet $l_1$ in $O$. Through $O$ draw $l_2$ in $N$ and
perpendicular to $l_1$.

Then the angle formed by the lines $l$ and $l_2$ is the plane angle
of the dihedral angle between the planes $M$ and $N$.
\hfill §~\Sref{106}
\begin{align*}
\therefore\quad l &\perp l_2  \tag*{§~\Sref{109}} \\
\therefore\quad l &\perp N.   \tag*{Why?}  \\
\tag*{\QED}
\end{align*}

\begin{Corollary}{115.}{}
If two planes are mutually perpendicular
and if a line is drawn from a point in one and perpendicular to
the other, then this line lies in the first plane and is perpendicular
to the intersection of the two planes.
\end{Corollary}

\Proof: By the theorem there is a line through $P$ which lies
in the plane $M$ and is $\perp N$. But through $P$ there is only one
line $\perp N$.\hfill    §~\Sref{80}

\NewFolio %% -----File: 039.png---Folio 31-------

\Section{PLANE PERPENDICULAR TO EACH OF TWO PLANES}

\begin{Theorem}{116.}{XVIII.}
If a plane is perpendicular
to each of two planes, it is perpendicular to their line
of intersection.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{039a}{0.6\textwidth}
\end{center}

\Objective{Given $Q\perp M$ and $Q\perp N$ and $l$ the intersection of $M$ and~$N$.}
{To prove that $Q\perp l$.}

\Proof: From a point $P$ common to $M$ and $N$ draw a line
$l'\perp Q$.

Then $l'$ lies in both $M$ and $N$.\hfill §~\Sref{115}

Hence, $l'$ and $l$ are the same line.\hfill Why?

That is, $l\perp Q$, or $Q\perp l$. \QED


\begin{Sightwork}

\item[1.] State theorems on dihedral angles corresponding to those in §§~\Sref{24},
\Sref{30}, \Sref{31}, \Sref{32}, on plane angles.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\item[2.] Name all dihedral angles in the accompanying
figure.

\item[3.] Any plane $\perp$ the edge of a dihedral angle is
$\perp$ each of its faces. Why?

\item[4.] If each of three lines is $\perp$ the other two at
the same point, then each is $\perp$ the plane of the
other two. Why?
\end{minipage}%
% [** Illustration]
\hfill\Input{039b}{1.25in}

\item[5.] Find the locus of all points at a given distance from a given plane
and also at a given distance from a second plane. Discuss this locus for
the various cases possible.

\end{Sightwork}

\NewFolio %% -----File: 040.png---Folio 32-------

\Section{PERPENDICULAR  LINES  AND  PLANES}

\begin{Theorem}{117.}{XIX.}
If a line is perpendicular to
a plane, every plane containing this line is perpendicular
to that plane.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{040a}{0.4\textwidth}
\end{center}

\Objective{Given a line $l\perp$ plane $N$ at $P$.}
{To prove that a plane $M$ containing $l$ is $\perp N$.}

\Proof: Let $l_1$ be the intersection of $M$ and $N$. In $N$ draw
$l_2\perp l_1$ at $P$.

Then\hfill $l\perp l_1$, $l_2\perp l_1$, and $l\perp l_2$.\hfill Why? \\
\null\hfill $\therefore\ M\perp N$.\hfill Why? \\
\QED

\begin{wrapfigure}[5]{r}{1.5in}
% [** illustration]
\Input{040b}{1.5in}
\end{wrapfigure}
\Paragraph{118.}{Projection of a Figure on a Plane.} The projection of a
\index{Foot@\textbf{Foot}!of a perpendicular}%{\Pnum{118}}%
\index{Plane@\textbf{Plane}!projections upon}%{\Pnum{118}}%
\index{Projection@\textbf{Projection}!of a figure}%{\Pnum{118}}%
point on a plane is the foot of the perpendicular from the
point to the plane. The projection  of
any \emph{figure} on a plane is the locus of the
projections of all points of the figure on
the plane.

\begin{Remark}
Thus, $B$ is the projection of the point $A$ on $M$,
and $l_2$ is the projection of the line $l_1$ on $M$.
\end{Remark}

\Paragraph{119.}{Angle Between Line and Plane.} The angle between a
% [** PP: Index entry points to \S 118]
\index{Angle@\textbf{Angle}!between a line and a plane}%{\Pnum{119}}%
\index{Angle@\textbf{Angle}!of projection}%{\Pnum{119}}%
\index{Projection angle@\textbf{Projection angle}}%{\Pnum{119}}%
plane and a line oblique to it is understood to be the \emph{acute
angle} formed by the line and its projection upon the plane.

\Paragraph{120.}{Bisector of a Dihedral Angle.} A half-plane bisects a
\index{Dihedral@\textbf{Dihedral angles}!bisector of}%{\Pnum{120}}%
dihedral angle if it passes through the edge of the dihedral
angle and bisects its plane angle.

\NewFolio %% -----File: 041.png---Folio 33-------

\Section{PROJECTION  OF  A  STRAIGHT  LINE ON  A  PLANE}

\begin{Theorem}{121.}{XX.}
\index{Projection@\textbf{Projection}!of a line-segment}%{\Pnum{121}}%
The projection of a straight
line on a plane is a straight line in that plane.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{041a}{0.4\textwidth}
\end{center}

\Objective{Given a plane $M$ and a line $AB$ projected upon it.}
{To prove that the projection is a straight line.}

\Proof: Project any three points $A$, $E$, $B$ of the line $AB$
into the points $C$, $F$, $D$ in the plane $M$.

Then $AC$ and $BD$ determine a plane $\perp M$.\hfill §§~\Sref{83}, \Sref{117}

Hence, by §§~\Sref{115}, \Sref{69}, any three points, $C$, $F$, $D$, of the projection
lie in a straight line, that is, the projection is a straight
line.
\QED % [** PP: Move \QED to same line]

\begin{Theorem}{122.}{XXI.}
The acute angle formed by a
straight line with its own projection on a plane is the
least angle which it makes with any line in that plane.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{041b}{0.4\textwidth}
\end{center}

\Objective{Given $CB$ the projection of $AB$ on the plane $M$, and any other
line $BD$ in $M$ through $B$.}
{To prove that $\angle ABD > \angle ABC$.}

\Proof: Draw $AC\perp M$, make $BD = BC$, and draw $AD$.
Then prove $AD > AC$ and $\therefore \angle ABD > \angle ABC$.\hfill §~\Sref{33}

\NewFolio %% -----File: 042.png---Folio 34-------

\Section{PLANE BISECTING A DIHEDRAL ANGLE}

\begin{Theorem}{123.}{XXII.}
The locus of all points equally
distant from the faces of a dihedral angle is the half-plane
bisecting the angle.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{042ab}{}
\end{center}

\Objective{Given the half-plane $P$ bisecting the dihedral $\angle M-AB-N$.}
{To prove {\upshape(1)} that any point $E$ in $P$ is equally distant from $M$
and~$N$, and {\upshape(2)} that any point which is equally distant from $M$
and $N$ lies in~$P$.}

\Proof: (1) Draw $EC\perp M$ and $ED\perp N$.

Then (a) Plane $CED\perp M$ and also $\perp N$.\hfill Why?

$\phantom{\text{Then}}$ (b) Plane $CED\perp AB$.\hfill Why?

Now $\angle COE$ is the plane $\angle$ of $M-AB-P$, \\
and $\angle DOE$ is the plane $\angle$ of $N-AB-P$.\hfill Why? \\
\null\hfill $\therefore \angle COE = \angle DOE$ by the hypothesis.\hfill \\
Hence\hfill $EC = ED$.\hfill §§~\Sref{42}, \Sref{23}

(2) Let $E$ be any point such that $EC = ED$, these being
perpendicular respectively to $M$ and $N$.

Then $\angle COE = \angle DOE$.\hfill §§~\Sref{40}, \Sref{23}

Hence, the plane $P$ determined by $ABE$ is the bisector of
the given dihedral angle, that is, the point $E$ lies in $P$.
\QED

\begin{Sightwork}

\item[] Define a locus. Why are two conditions necessary in determining a
locus? Are these conditions fulfilled in the above theorem?
\end{Sightwork}

\NewFolio %% -----File: 043.png---Folio 35-------

\Section{PROBLEMS IN CONSTRUCTION}
\SecToc{Constructions of Planes and Lines}

In making constructions in solid geometry, it is assumed that

\index{Axioms@\textbf{Axioms}|(}%{\Pnum{124}}%
\textbf{124.} \textit{A line can be passed through any two points.}

\textbf{125.} \textit{A plane can be passed through three non-collinear points.}

\textbf{126.} \textit{Through a point not on a given line there can be drawn
a line parallel to this line.}
\Unit{126.}%

\textbf{127.} \textit{Through a point a line perpendicular to a given line can
be drawn.}
\Unit{127.}%

\begin{Corollary}{128.}{}
A plane can be passed through \Inum{(1)}~a line
and a point not on it, \Inum{(2)}~two intersecting lines, \Inum{(3)}~two parallel
lines.
\end{Corollary}
\hfill §~\Sref{73} \\
\QEF
\index{Axioms@\textbf{Axioms}|)}%{\Pnum{128}}%


\Section{CONSTRUCTING A PLANE PERPENDICULAR TO A LINE}

\begin{Problem}{129.}
Through a given point to construct
a plane perpendicular to a given line.
\end{Problem}

% [** Illustrations] FIG. 1.    FIG. 2.
\begin{center}
\Input{043ab}{}
\end{center}

\Objective{\Inum{(1)}~Given a line~$l$ and a point~$P$ on it. \textsc{Fig.~1.}}
{To construct a plane $M \perp l$ at~$P$.}

\begin{Construction}
Through $P$ draw two lines $PA$ and $PB$ each
perpendicular to~$l$.\hfill §§~\Sref{127},~\Sref{75}

Then the plane $M$ determined by $PA$ and~$PB$ is $\perp l$.\hfill §~\Sref{76}
\end{Construction}

\Objective{\Inum{(2)}~Given a line~$l$ and a point~$P$ not on it. \textsc{Fig.~2.}}
{To construct a plane~$N$ through~$P$ and $\perp l$.}

\begin{Construction}
From $P$ draw a line $PQ \perp l$.\hfill §~\Sref{127}

At $Q$ construct a plane $N \perp l$, as in~(1).

Then, by §~\Sref{79}, $PQ$ lies in $N$, and hence~$N$ is the plane required. \\
\QEF
\end{Construction}

\NewFolio %% -----File: 044.png---Folio 36-------

\Section{CONSTRUCTING A LINE PERPENDICULAR TO A PLANE}

\begin{Problem}{130.}
Through a given point to construct a
line perpendicular to a given plane.
\end{Problem}

% [** Illustrations]
\begin{center}
\Input{044ab}{}
\end{center}

\Objective{\Inum{(1)}~Given a plane $M$ and a point $P$ in it.}
{To construct a line through $P$ and $\perp M$.}

\begin{Construction}
Let $l$ be any line in $M$ not passing through $P$.

Draw $PB \perp l$. Through $B$ draw $BK \perp l$ but not in~$M$.

In the plane $PBK$ draw $PC \perp PB$ and meeting $BK$ in~$C$.

Prove $PC \perp M$.\hfill §§~\Sref{76}, \Sref{117}, \Sref{114}
\end{Construction}

\Objective{\Inum{(2)}~Given a plane $N$ and a point $P$ not in it.}
{To construct a line through $P$ and $\perp N$.}

\begin{Construction}
Let $l$ be any line in $N$. Draw $PB \perp l$ and in
$N$ draw $BK \perp l$. In the plane $PBK$ draw $PC \perp BK$.

Prove, as in~(1), $PC \perp N$.\hfill\hbox{(See also §~\Sref{80}.)} \\
\QEF
\end{Construction}

\Section{CONSTRUCTING A PLANE PARALLEL TO A LINE}

\begin{Problem}{131.}
Through one of two skew lines to
pass a plane parallel to the other.
\end{Problem}

\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Objective{Given two skew lines $l_1$ and $l_2$.}
{To construct a plane through $l_1 \parallel l_2$.}
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{044c}{1in}}

\begin{Construction}
Through a point $P$ in $l_1$ pass a
line $l_3 \parallel l_2$.\hfill\hbox{(§~\Sref{126}.)}

The plane determined by $l_1$ and $l_3$ is $\parallel l_2$.\hfill §~\Sref{90} \\
\QEF
\end{Construction}

\NewFolio %% -----File: 045.png---Folio 37-------

\Section{CONSTRUCTING A PLANE PERPENDICULAR TO A PLANE}

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\begin{Problem}{132.}
Through a given line to pass a
plane perpendicular to a
given plane.
\end{Problem}

\Objective{Given a line $l$ and a plane $M$.}
{To pass a plane through $l$ and
$\perp M$.}
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{045a}{2in}}

\begin{Construction}
Through a point $P$ in $l$ pass a line $l_1 \perp M$.
\hfill §~\Sref{130}

The plane $N$ determined by $l$ and $l_1$ passes through $l$ and
is $\perp M$.\\
\null\hfill §~\Sref{117} \\
\QEF
\end{Construction}


\Section{CONSTRUCTING A PLANE PARALLEL TO EACH OF TWO LINES}

\begin{Problem}{133.}
Through a point to pass a plane
parallel to each of two given lines.
\end{Problem}

\Objective{Given a point $P$ and two lines $l_1$ and~$l_2$.}
{To construct a plane through $P$ parallel to $l_1$ and~$l_2$.}

\begin{Construction}
Through $P$ pass lines $l'_1$ and $l'_2$ such that
$l'_1 \parallel l_1$ and $l'_2 \parallel l_2$.\hfill §~\Sref{126}

Then the plane determined by $l'_1$ and $l'_2$ is parallel to both
$l_1$ and~$l_2$. \\
\null\hfill §~\Sref{90} \\
\QEF
\end{Construction}


% SIGHT WORK
\begin{Sightwork}

\item[1.] Study the special case of §~\Sref{132} in which the given line is perpendicular
to the given plane.

\item[2.] Is it possible that the plane constructed by the method of §~\Sref{133} may
contain one or both of the given lines? Study all possible cases.

\item[3.] Through a given point pass a plane $\perp$ each of two given planes.
How many such can be constructed? Study the case in which the two
given planes are parallel to each other.

\end{Sightwork}

\begin{Suggestion}
Consider the relation of the required plane to the intersection
of the two given planes.
\end{Suggestion}

\NewFolio %% -----File: 046.png---Folio 38-------

\Section{COMMON PERPENDICULAR TO TWO SKEW LINES}

\begin{Problem}{134.}
To construct a common perpendicular
to two skew lines.
\end{Problem}

% [** Illustration]
\vspace*{-0.375in}
\begin{center}
\Input{046a}{0.5\textwidth}
\end{center}
\vspace*{-0.25in}

\Objective{Given two skew lines $l_1$ and $l_2$.}
{To construct a line $BC$ perpendicular to each of them.}

\begin{Construction}
Through $A$, any point in $l_2$, draw $l_3 \parallel l_1$.

Let $M$ be the plane determined by $l_2$ and $l_3$.

Through $l_1$ pass a plane $N \perp M$ and meeting $M$ in $l_4$.

At $B$ the intersection of $l_2$ and $l_4$ erect $BC \perp M$.

Then $BC$ is the required line perpendicular to $l_1$ and $l_2$.
\end{Construction}

\textbf{Outline of Proof}: Show each of the following:

(1)~$BC$ lies in the plane $N$, and $\therefore$ meets $l_1$.

(2)~$l_1 \parallel l_4$, (3)~$BC \parallel l_4$, and $\therefore BC \perp l_1$.

\begin{Corollary}{135.}{}
There cannot be more than one common
perpendicular to two skew lines.
\end{Corollary}

\begin{Suggestion}
Suppose a second common perpendicular drawn from a
point $P_1$ in $l_1$ to a point $P_2$ in $l_2$. Then show that from $P_1$ it would be
possible to have two lines $\perp$ plane $M$.
\end{Suggestion}

\begin{Sightwork}

\item[1.] If a line is $\perp$ a plane, show that its projection is a point.

\item[2.] If a line-segment is $\parallel$ a plane, show that its projection is a segment
equal to the given segment.

\item[3.] If a line-segment is oblique to a plane, show that its projection is
less than the given segment.
\end{Sightwork}

\NewFolio %% -----File: 047.png---Folio 39-------

\begin{Exercises}

\item[1.] Show that the projections of two parallel lines on a plane
are parallel. Is the converse true? Illustrate with pieces of
cardboard.

\item[2.] If two parallel lines meet a plane, they make equal
angles with it. Why? Is the converse true? What is the
corresponding theorem in plane geometry? Is its converse
true?

\item[3.] If a line cuts two parallel planes, it makes equal angles
with them. Why? Is the converse true? Discuss the corresponding
theorems in plane geometry.

\item[4.] If two parallel line-segments are oblique to a plane,
their projections on the plane are in the same ratio as the given
segments.

\item[5.] A line and its projection on a plane determine a plane
perpendicular to the given plane.

\item[6.] If a line is parallel to one of two given planes and perpendicular
to the other, then the two planes are perpendicular
to each other.

\item[7.] Find the locus of all points equidistant from two points
$A$ and~$B$ and also at a given distance from a plane~$M$.
Discuss.\index{Loci@\textbf{Loci problems}|(}%

\item[8.] Find the locus of all points equidistant from two given
planes and also equidistant from two given points. Discuss.

\item[9.] Find the locus of all points equidistant from two given
planes $M$ and~$N$, and also equidistant from two other given
planes $Q$ and~$R$. Discuss. Compare with the corresponding
loci in plane geometry.
\index{Loci@\textbf{Loci problems}|)}%

\item[10.] Prove that there is a line in space every point of which
is equidistant from three points $A$, $B$, $C$, provided these points
do not lie on one line.
\end{Exercises}

\NewFolio %% -----File: 048.png---Folio 40-------

\Section{POLYHEDRAL ANGLES}
\SecToc{Polyhedral Angles}

\Paragraph{136.}{Polyhedral Angle.} Given a convex polygon and a point
\index{Angle@\textbf{Angle}!polyhedral}%{\Pnum{136}}%
\index{Polyhedral@\textbf{Polyhedral angles}}%{\Pnum{136}}%
$P$ not in its plane. If a half-line~$l$ with its end point fixed at
$P$ moves so that it always touches the polygon and is made to
traverse it completely, it is said to generate a \emph{convex polyhedral
angle}.

\begin{wrapfigure}[7]{r}{1.5in}
% [** Illustration]
\Input{048a}{1.5in}
\end{wrapfigure}
\Paragraph{137.}{Vertex. Edges.} The fixed point is the \emph{vertex} of the
\index{Edge@\textbf{Edge}!of a polyhedral angle}%{\Pnum{137}}%
\index{Faces@\textbf{Faces}!of a polyhedral angle}%{\Pnum{137}}%
\index{Polyhedral@\textbf{Polyhedral angles}!edges of}%{\Pnum{137}}%
\index{Vertex@\textbf{Vertex}!of a polyhedral angle}%{\Pnum{137}}%
polyhedral angle, and the rays through
the vertices of the polygon are the \emph{edges}
of the polyhedral angle.

Any two consecutive edges determine a
plane, and the portion of such a plane
included between these edges is called a
\emph{face} of the polyhedral angle.

\Paragraph{138.}{Face Angles.} The plane angles at the vertex are called
\index{Angle@\textbf{Angle}!face}%{\Pnum{138}}%
the \emph{face angles} of the polyhedral angle. A polyhedral angle
having three faces is called a \emph{trihedral angle}.
\index{Angle@\textbf{Angle}!trihedral}%{\Pnum{138}}%
\index{Polyhedral@\textbf{Polyhedral angles}!faces of}%{\Pnum{138}}%
\index{Trihedral@\textbf{Trihedral angles}}%{\Pnum{138}}%

\begin{Remark}
Thus in the figure, $P$ is the vertex, $PA$, $PB$, etc., are the edges, and
$\angle APB$, $\angle BPC$, etc., are the face angles.
\end{Remark}

\Paragraph{139.}{Equal Polyhedral Angles.} Two polyhedral angles are
\emph{equal} if they can be made to coincide.
\index{Polyhedral@\textbf{Polyhedral angles}!equal}%{\Pnum{139}}%

\begin{Remark}
A polyhedral angle is read by naming the vertex and one letter in
each edge, as $P-ABCDE$, or by naming the vertex alone where no ambiguity
would arise.
\end{Remark}

\Paragraph{140.}{Order of Parts in Triangles.} In  the triangles $ABC$,
\begin{wrapfigure}[5]{r}{3in}
% [** Illustration]
\Input{048bc}{3in}
\end{wrapfigure}
\index{Order@\textbf{Order of parts}!in a triangle}%{\Pnum{140}}%
and $A'B'C'$, $AB = A'B'$,
$BC = B'C'$,  $CA = C'A'$.
However, the sides $AB$,
$BC$, $CA$, are arranged
in counter-clockwise order, and the sides $A'B'$, $B'C'$, $C'A'$, are
arranged in clockwise order. That is, the parts of the two
triangles are arranged in opposite orders.

\NewFolio %% -----File: 049.png---Folio 41-------

\noindent\begin{minipage}[b]{\textwidth-1.875in}
\Indent\Paragraph{141.}{Order of Parts in Polyhedral Angles.}
\index{Order@\textbf{Order of parts}!in a polyhedral angle}%{\Pnum{141}}%
In the trihedral angles $O$
and $O'$ the face angles $AOB$, $BOC$,
$COA$ are arranged in \emph{counter-clockwise}
order as viewed from the vertex,
while  $A'O'B'$, $B'O'C'$, $C'O'A'$ are
arranged in \emph{clockwise} order.
\end{minipage}%
% [** Illustration]
\hfill\Input{049ab}{1.875in}

\begin{Remark}
The trihedral angles $O$ and $O'$ cannot be made to coincide, even though
their corresponding parts are equal. This can be illustrated by attempting
to put a left glove on the right hand.
\end{Remark}

\Section{CONDITIONS WHICH MAKE TRIHEDRAL ANGLES EQUAL}

\begin{Theorem}{142.}{XXIII.}
Two trihedral angles are
equal if two face angles and the included dihedral
angle of one are equal respectively to two face angles
and the included dihedral angle of the other and arranged
in the same order.
\end{Theorem}

% [** Illustration]
\vspace*{-0.25in}
\begin{center}
\Input{049cd}{}
\end{center}
\vspace*{-0.25in}

\textbf{Suggestion for Proof}: This and the following theorem may be
proved by superposition in the same manner as the corresponding
theorems on the equality of triangles in plane geometry.

\begin{Theorem}{143.}{XXIV.} Two trihedral angles are
equal if a face angle and the two adjacent dihedral
angles of one are equal respectively to a face angle and
the adjacent dihedral angles of the other and arranged
in the same order.
\end{Theorem}

\NewFolio %% -----File: 050.png---Folio 42-------

\Section{EQUAL DIHEDRAL ANGLES OPPOSITE EQUAL FACE ANGLES}

\begin{Theorem}{144.}{XXV.}
If two trihedral angles have
the three face angles of the one equal respectively to the
three face angles of the other, the dihedral angles opposite
the equal face angles are equal.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{050ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given the trihedral angles $P$ and $P'$, in which $\angle a = \angle a'$, $\angle b
= \angle b'$, $\angle c = \angle c'$.}
{To prove that the corresponding dihedral angles are equal.}

\textbf{Outline of Proof}: Let $P$ and $P'$ be cut by the planes $ABC$ and
$A'B'C'$, making $PA = PB = PC = P'A' = P'B' = P'C'$.

On the edges $PA$ and $P'A'$ lay off $AF = A'F'$, and through
$F$ and $F'$ pass planes $\perp$ to $PA$ and $P'A'$ respectively.

\begin{Remark}
$AF$ and $A'F'$ are to be taken short enough so that these planes shall
cut the segments $AB$, $AC$ and $A'B'$, $A'C'$ respectively.
\end{Remark}

In order to prove that $\angle B-AP-C = \angle B'-A'P'-C'$, i.e.~that
$\angle DFE = \angle D'F'E'$, use the pairs of $\triangles APB$, $A'P'B'$; $BPC$,
$B'P'C'$; $CPA$, $C'P'A'$; $ABC$, $A'B'C'$.    Then prove in order:
\begin{gather*}
(1)\quad \triangle ADF = \triangle A'D'F'; \qquad (2)\quad \triangle AEF = \triangle A'E'F'; \\
(3)\quad \triangle ADE = \triangle A'D'E'; \qquad (4)\quad \triangle DEF = \triangle D'E'F'. \\
\therefore\quad \angle DFE = \angle D'F'E'.
\end{gather*}

\begin{Corollary}{145.}{}
Two trihedral angles are equal if the face
angles of one are equal respectively to the face angles of the other
and arranged in the same order.
\end{Corollary}

\NewFolio %% -----File: 051.png---Folio 43-------

\Section{SYMMETRICAL TRIHEDRAL ANGLES}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{146.}{Symmetrical Trihedral Angles.} Two
\index{Polyhedral@\textbf{Polyhedral angles}!symmetrical}%{\Pnum{146}}%
\index{Symmetrical@\textbf{Symmetrical}!trihedral angles}%{\Pnum{146}}%
\index{Trihedral@\textbf{Trihedral angles}!symmetrical}%{\Pnum{146}}%
trihedral angles are \emph{symmetrical}, one to the
other, if the face angles and the dihedral
angles of one are equal respectively to the
face angles and the dihedral angles of the
other, but arranged in the \emph{opposite order}.

\Indent\Paragraph{147.}{Vertical Trihedral Angles.} Two trihedral
\index{Trihedral@\textbf{Trihedral angles}!vertical}%{\Pnum{147}}%
angles are \emph{vertical} if the edges of
one are the prolongations of the edges of
the other.
\end{minipage}%
% [** Illustration]
\hfill\Input{051a}{1.25in}


\begin{Corollary}{148.}{1.}
Two vertical trihedral angles are symmetrical.
\end{Corollary}

\begin{Remark}
The proof is evident from the figure above.
\end{Remark}

\begin{Corollary}{149.}{2.}
Two trihedral angles are symmetrical if
the face angles of one are equal to the face angles of the other and
arranged in the opposite order.
\end{Corollary}

\begin{Sightwork}

\item[1.] Is it possible that all three face angles of a trihedral angle shall be
right angles? Is it possible that all three dihedral angles shall be right
angles?

\begin{Suggestion}
Consider a corner of a rectangular box.
\end{Suggestion}

\item[2.] If three planes meet in a point how many trihedral angles are
formed? If all face angles of one of these trihedral angles are right
angles, what about the face angles of the other trihedral angles?

\item[3.] If in the figure of §~\Sref{144} the $\angle a = \angle b = \angle c = 60$°, what can be said
about the triangles $PAB$, $PBC$, $PCA$, and $ABC$?

\item[4.] Show how to locate a point which is at a distance of $2$~feet from
each of the three planes determined by the faces of a trihedral angle.
How many such points are there?

\begin{Suggestion}
Pass planes parallel to each face of the trihedral angle,
one on each side of it, and at a distance of $2$~feet from it.
\end{Suggestion}
\end{Sightwork}

\NewFolio %% -----File: 052.png---Folio 44-------

\Section{CONDITIONS WHICH MAKE TRIHEDRAL ANGLES  SYMMETRICAL}

\begin{Theorem}{150.}{XXVI.}
Two trihedral angles are
symmetrical
\end{Theorem}

\Inum{(1)}~if two face angles and the included dihedral
angle of one are equal respectively to two face angles
and the included dihedral angle of the other, but arranged
in the opposite order; or

\Inum{(2)}~if a face angle and the adjacent dihedral angles
of one are equal respectively to a face angle and the
adjacent dihedral angles of the other, but arranged in
the opposite order.

% [** Illustration]
\begin{center}
\Input{052abc}{}
\end{center}

\Proof: Let $t_1$ and $t_2$ be two trihedral angles having the
properties specified under~(1). Let $t_3$ be a trihedral angle
symmetrical to $t_1$. Then $t_2 = t_3$.\hfill §~\Sref{142}
\[
\therefore\quad\text{$t_1$ and $t_2$ are symmetrical.}
\]

The proof of the second part is left to the student.

\begin{Theorem}{151.}{XXVII.}
The sum of two face angles
of a trihedral angle is greater than the third face angle.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\textbf{Outline of Proof}: Connect points $A$ and $C$ on two
sides. Suppose $\angle ADC > \angle ADB$. Construct $DE$
making $\angle ADE = \angle ADB$. Suppose that point $B$ is
taken so that $DB = DE$, and that $AB$ and $BC$ are
drawn. Then prove
\end{minipage}%
% [** Illustration]
\hfill\raisebox{-0.25in}{\smash{\Input{052d}{1.25in}}}
\Indent\Indent\begin{minipage}{3.5in}
\noindent%
(1)\quad $AB = AE$,\qquad (2)\quad $AC < AB + BC$, \\
(3)\quad $EC < BC$,\qquad (4)\quad $\angle EDC < \angle BDC$, \\
(5)\quad $\angle ADC < \angle ADB + \angle BDC$.
\end{minipage}%

\NewFolio %% -----File: 053.png---Folio 45-------

\Section{SUM OF THE FACE ANGLES OF A POLYHEDRAL ANGLE}

\begin{Theorem}{152.}{XXVIII.}
The sum of the face angles
of any convex polyhedral angle is less than four right
angles.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{053a}{0.5\textwidth}
\end{center}

\Proof: Let $ABCDE$ be a plane section of the given polyhedral
angle. The number of triangles thus formed having
$P$ for a vertex is equal to the number of face angles of the
polyhedral angle.

Let $O$ be any point in the base, and draw $OA$, $OB$, $OC$, etc.

Then $\angle PBA + \angle PBC > \angle ABC$, $\angle PCB + \angle PCD >
\angle BCD$, and so on.\hfill §~\Sref{151}

Now the sum of the $\Angles$ of the $\triangles OAB$, $OBC$, etc., is equal to
the sum of the $\Angles$ of the $\triangles PAB$, $PBC$, etc.

Hence, $\angle APB + \angle BPC + \dotsb < \angle AOB + \angle BOC + \dotsb$.

But the sum of the $\Angles$ about $O$ is four right angles.

Therefore, the sum of the face angles of the polyhedral
angle is less than four right angles.
\QED

\textit{Note}. If as in §~\Sref{144} the three edges of a trihedral angle are cut by a
plane, the intersection is a triangle. Then the sides and angles of such
a triangle \emph{correspond} to the face angles and dihedral angles respectively
of the trihedral angle.

\begin{Sightwork}

\item[1.] What theorem in plane geometry corresponds to the theorem of
§~\Sref{151} in the sense of the above note?

\item[2.] Discuss propositions of plane geometry corresponding to those
of §~\Sref{150}.
\end{Sightwork}

\NewFolio %% -----File: 054.png---Folio 46-------

\Section{SUMMARY OF BOOK I}
\index{Summaries@\textbf{Summaries}}

\begin{Itemize}

\item[1.] State the axioms used in this Book.

\item[2.] Describe the various ways of determining a plane.

\item[3.] State the definitions on perpendicular lines and planes.

\item[4.] State a theorem on a line through a point perpendicular
to a plane.

\item[5.] State a theorem on a plane through a point perpendicular
to a line.

\item[6.] State the definitions on parallel lines and planes.

\item[7.] State propositions on (1)~a plane through a point parallel
to a given plane; (2)~a plane through a line parallel to a given
line; (3)~a plane through a point parallel to two given lines.

\item[8.] State some facts about perpendiculars in the plane
which do not hold in space.

\item[9.] Define a dihedral angle and the plane angle of a dihedral
angle.

\item[10.] What theorems on perpendicular planes are proved in
connection with dihedral angles? Notice that these theorems
include all those on one plane perpendicular to another. Why
should this be the case?

\item[11.] State the definitions and theorems on projections given
in this Book.

\item[12.] State the definitions and theorems on polyhedral angles
thus far given.

\item[13.] Define \emph{symmetrical polyhedral angles}. Are symmetrical
trihedral angles equal?

\item[14.] State theorems on equal and symmetrical trihedral
angles.

\item[15.] Give examples of loci in plane geometry which differ from
corresponding loci in solid geometry. See, for instance, §~\Sref{6}.
\end{Itemize}

\NewFolio %% -----File: 055.png---Folio 47-------

\Section{MISCELLANEOUS EXERCISES ON BOOK I}

\begin{Itemize}

\item[1.] A Christmas tree is made to stand on a cross-shaped
base. What must be the relation of the tree to each piece of
the cross in order that it may be perpendicular to the floor?

\item[2.] If $A$, $B$, and~$C$ do not lie in the same line, and if their
projections on a plane~$M$ do lie in a straight line, what is the
relation of the planes $M$ and~$ABC$?

\item[3.] Is it possible to project a circle upon a plane so that the
projection shall be a straight line-segment? If so, how must
the circle and the plane be related?

\item[4.] If the projections of a set of points on each of two planes
not parallel to each other lie in straight lines, show that the
points themselves lie in a straight line.

\item[5.] Find the locus of points $3$~feet from one of two intersecting
planes and $6$~feet from the other.
\index{Loci@\textbf{Loci problems}|(}%

\item[6.] It is required that a series of electric lights shall be
$7$~feet above the floor of a room and $3$~feet from the walls.
Find the locus of all points at which such lights may be placed.

\item[7.] Find the locus of a point in space such that the difference
of the squares of its distances from two fixed points, $A$ and $B$,
is constant.

\begin{Suggestion}
Through either $A$ or~$B$ construct a plane $\perp AB$.
\end{Suggestion}

\item[8.] Given two points $A$ and~$B$ on the same side of a
plane~$M$. Determine a point~$P$ in $M$ such that $AP + PB$
shall be a minimum.

\begin{Suggestion}
Pass a plane through $A$ and~$B$ perpendicular to~$M$, and
proceed as in Ex.~10, page~280, Plane Geometry.
\end{Suggestion}

\item[9.] Show that if the edge of a dihedral angle is cut by two
parallel planes, the sections which they make with the faces
form equal angles. Is the converse proposition true?

\item[10.] Show that if all edges of a trihedral angle are cut by a
series of parallel planes, the intersections form a series of
similar triangles. Is the converse proposition true?
\end{Itemize}

\NewFolio %% -----File: 056.png---Folio 48-------

\begin{Itemize}
\item[11.] Find the locus of the intersection points of the medians
of the triangles obtained in Ex.~10. Also of the altitudes.

\item[12.] How many planes may be made to pass through a given
point parallel to a given line? Discuss the mutual relation of
all such planes.

\item[13.] Through a point~$P$ construct a line meeting each of two
lines $l_1$ and $l_2$. %[** PP: Missing end-of-sentence period added]

\begin{Suggestion}
Let $M$ and $N$ be the planes determined by $P$ and~$l_1$, $P$
and~$l_2$. Consider their intersection. Is this construction always possible?
\end{Suggestion}

\item[14.] A line $l$ is parallel to a plane $M$, and lines $l_1$ and~$l_2$ in~$M$
are not parallel to~$l$. Show that the shortest distance between
$l$ and~$l_1$ is equal to the shortest distance between $l$ and~$l_2$.

\item[15.] Find the locus of all points equidistant from the planes
determined by the faces of a trihedral angle.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\item[16.] Find the locus of all points equidistant
from the edges of a trihedral angle.
\index{Loci@\textbf{Loci problems}|)}%

\begin{Suggestion}
On the edges lay off $PA = PB = PC$.
Let $O$ be equidistant from $A$, $B$, and~$C$. Then
any point $Q$ in $PO$ is equidistant from the edges.
\end{Suggestion}

\item[17.] Planes determined  by the edges of
a trihedral angle and the bisectors of the
opposite face angles meet in a line.
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.5in}{\smash{\Input{056a}{1.5in}}}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\item[18.] If three planes are so related that the segments intercepted
on any transversal line are in the
same ratio as the segments intercepted on
any other transversal then the planes are
parallel.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{056b}{1.55in}}

\begin{Suggestion}
Let $M$, $N$, $Q$ be the three planes.
From $A$ any point in $M$ draw three lines, not in the
same plane, meeting $N$ in $A'$, $B'$, $C'$, and $Q$ in
$A''$, $B''$, $C''$. Use the hypothesis to show that $A'B' \parallel A''B''$ and
$C'B' \parallel C''B''$. Hence $Q \parallel N$. Similarly show that $M \parallel N$.
\end{Suggestion}

\end{Itemize}

\NewFolio %% -----File: 057.png---Folio 49-------

\begin{BioSketch}{Thales}{Thales of Miletus}
  (640--542~B.C\@.) %[** PP: Added missing period, was B C.]
  was one of the Seven Wise Men of
  Greece. He learned astronomy and geometry in Egypt and was the first
  to introduce the scientific study of geometry in Greece. He measured
  the height of the Pyramids in Egypt by similar triangles and found a
  method of computing the distance of a ship at sea. He predicted the
  solar eclipse of 585~B.C\@.
\end{BioSketch}

\NewFolio %% -----File: 058.png---Folio 50-------

\Chapter{BOOK II}{Regular Polyhedrons}
% [** PP: Explicit pagination to accommodate large illo on next page]
\enlargethispage{36pt}

\Paragraph{153.}{Polyhedron.} A \emph{polyhedron} % [** PP: Emphasis added]
\index{Polyhedrons@\textbf{Polyhedrons}}%{\Pnum{153}}%
is a geometric solid whose
boundary consists of plane polygons.

\Paragraph{154.}{Convex Polyhedron.} A polyhedron
is \emph{convex} if every section of it made by a
plane is a convex polygon.
\index{Polyhedrons@\textbf{Polyhedrons}!convex}%{\Pnum{154}}%

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{155.}{Faces, Vertices, Edges.} The polygons
\index{Edge@\textbf{Edge}!of a polyhedron}%{\Pnum{155}}%
\index{Faces@\textbf{Faces}!of a polyhedron}%{\Pnum{155}}%
\index{Polyhedrons@\textbf{Polyhedrons}!edges of}%{\Pnum{155}}%
\index{Polyhedrons@\textbf{Polyhedrons}!faces of}%{\Pnum{155}}%
\index{Polyhedrons@\textbf{Polyhedrons}!vertices of}%{\Pnum{155}}%
\index{Vertices@\textbf{Vertices}!of a polyhedron}%{\Pnum{155}}%
which bound the polyhedron are its \emph{faces};
the sides of these faces are the \emph{edges} and their vertices are the
\emph{vertices} of the polyhedron.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{058a}{1.25in}}

\Paragraph{156.}{Surface of Polyhedron.} The faces, edges, and vertices
taken together form the surface of the polyhedron.
\index{Polyhedrons@\textbf{Polyhedrons}!surface of}%{\Pnum{156}}%
\index{Surface@\textbf{Surface}!of a polyhedron}%{\Pnum{156}}%

\Paragraph{157.}{Polyhedrons Classified According to the Number of Faces.}
\index{Dodecahedron@\textbf{Dodecahedron}}%{\Pnum{157}}%
\index{Icosahedron@\textbf{Icosahedron}}%{\Pnum{157}}%
\index{Octahedron@\textbf{Octahedron}}%{\Pnum{157}}%
\index{Polyhedrons@\textbf{Polyhedrons}!classified}%{\Pnum{157}}%
\index{Tetrahedron@\textbf{Tetrahedron}}%{\Pnum{157}}%
A polyhedron of four faces is a \emph{tetrahedron}, one of six faces is
a \emph{hexahedron}, one of eight faces an \emph{octahedron}, one of twelve
faces a \emph{dodecahedron}, and one of twenty faces an \emph{icosahedron}.

\begin{Remark}
These names are all derived from the Greek and refer to the number of
faces. Compare other words derived from the same Greek roots such as
hexagon, octagon, dodecagon, etc.
\end{Remark}

\begin{Sightwork}

\item[1.] What is the smallest number of polygons which maybe used to
inclose a portion of space?

\Indent\begin{Suggestion}
Remembering that a triangle is regarded
as a polygon, consider a triangular pyramid.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\item[2.] How many edges and how many vertices has a
tetrahedron?

\Indent\item[3.] How many faces, edges, and vertices has a hexahedron?
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{058b}{1.25in}}

\Erratum{err058a}%
\begin{Suggestion}
Consider a cube.\footnoteTrans{See \Tref{058a}{Transcriber's Notes}.}
%[** PP: Mathematical error]
\end{Suggestion}
\end{Sightwork}

\NewFolio %% -----File: 059.png---Folio 51-------

\Section{MODELS OF THE REGULAR POLYHEDRONS}

\Paragraph{158.}{Regular Polyhedrons.} A polyhedron whose polyhedral
\index{Dodecahedron@\textbf{Dodecahedron}}%{\Pnum{158}}%
\index{Polyhedrons@\textbf{Polyhedrons}!regular}%{\Pnum{158}}%
\index{Regular@\textbf{Regular}!polyhedrons}%{\Pnum{158}}%
\index{Tetrahedron@\textbf{Tetrahedron}!regular}%{\Pnum{158}}%
angles are equal and whose faces are equal regular polygons is
a \emph{regular polyhedron}. The following are regular polyhedrons:

% [** Illustrations]
% TETRAHEDRON HEXAHEDRON OCTAHEDRON DODECAHEDRON ICOSAHEDRON
\begin{center}
\Input{059a_e}{\textwidth}
\end{center}

\Paragraph{159.}{Models of the Regular Polyhedrons.} Models of the regular
\index{Polyhedrons@\textbf{Polyhedrons}!models of}%{\Pnum{159}}%
polyhedrons may be made with cardboard by cutting out patterns,
as shown in the figures below, folding along the dotted
lines, and fastening the edges together by means of gummed
paper strips corresponding to the dotted margins.

% [** Illustrations]
\Page{50}%
\begin{center}
\Input{059f_j}{\textwidth}
\end{center}

\NewFolio %% -----File: 060.png---Folio 52-------

\begin{Sightwork}

\Page{51}%
\item[1.] What is the smallest possible number of face angles in a polyhedral
angle?

\item[2.] How many faces meet in a vertex of a regular tetrahedron?
What is the sum of the face angles forming one of its polyhedral angles?

% [** Illustrations]
\begin{center}
\Input{060abc}{\textwidth}
\end{center}

\item[3.] How many faces meet in a vertex of a regular octahedron? in a
regular icosahedron?

\item[4.] What is the sum of the face angles forming a polyhedral angle in
a regular octahedron? in a regular icosahedron?

\item[5.] Why may not six or more equilateral triangles be used to form a
polyhedral angle?

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\item[6.] What is the sum of the face angles forming
a polyhedral angle of a cube (hexahedron)?

\Indent\item[7.] Why may not four or more squares be used
to form a polyhedral angle?

% [** Illustration]

\Indent\item[8.] How many faces meet in a vertex of a
regular dodecahedron? What is the sum of the
face angles forming one of its polyhedral angles?

\Indent\item[9.] Why may not more than three regular pentagons
be used to form a polyhedral angle?
\end{minipage}%
% [** Illustration]
\hfill\raisebox{-0.25in}{\smash{\Input{060de}{1.25in}}}

\Indent\item[10.] May regular hexagons be used to form a polyhedral
angle? Why?

\item[11.] May any regular polygons of more than six
sides be used to form polyhedral angles? Why?

\item[12.] Can we show that not more than three different
regular polyhedrons may be formed having
triangles as their faces?

\item[13.] How many regular polyhedrons may be formed having squares as
their faces? How many having pentagons as their faces?
\end{Sightwork}

\NewFolio %% -----File: 061.png---Folio 53-------

\Section{THE NUMBER OF REGULAR POLYHEDRONS}
\SecToc{Construction of Regular Polyhedrons}

\begin{Theorem}{160.}{I.}
There are exactly five  regular
polyhedrons.
\end{Theorem}

\Proof: On page~\Pgref{50} it was shown how models of five different
regular polyhedrons may be made.

Hence there are at least five such polyhedrons.

That there cannot be more than five regular polyhedrons
follows from the two propositions:

\begin{Remark}
(a)~\textit{Every polyhedral angle has at least three face angles}.

(b)~\textit{The sum of the face angles of a polyhedral angle is less than~$360$°}.

From these propositions it follows that each of the polyhedral angles of
a regular polyhedron may be formed by three, four, or five (but not six)
equilateral triangles; by three (but not four) squares; or by three (but
not four) regular pentagons. Regular polygons having more than five
sides cannot form a polyhedral angle.
\end{Remark}

The details of the proof are left to the student. The sight
work on page~\Pgref{51} will furnish suggestions.

\Paragraph{161.}{Construction of Regular Polyhedrons.} The regular polyhedrons
\index{Regular@\textbf{Regular}!polyhedrons}%{\Pnum{161}}%
\index{Regular@\textbf{Regular}!tetrahedrons}%{\Pnum{161}}%
may be constructed by use of §~\Sref{130}. The construction
for the tetrahedron and octahedron are given below.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent(1)~\textbf{The regular tetrahedron}.\quad At the center~$E$ of
an equilateral triangle~$ABC$ erect a perpendicular
to the plane of the triangle. On this take a point~$D$
so that $AD = AC$.

Then the four triangles, $ABC$, $ACD$, $ABD$, $BCD$,
are \emph{regular} and equal, and the four trihedral angles
are equal. Why?
\end{minipage}%
% [** Illustration]
\hfill\Input{061a}{1.25in}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent(2)~\textbf{The regular octahedron}.\quad Through the center
$O$ of a square~$ABCD$ draw a perpendicular to the
plane of the square and on it take points $E$ and~$F$
such that $AF = AE = AB$. Join $E$ and~$F$ to each
of the four vertices, $A$, $B$, $C$, $D$.

Then the eight faces are equal regular triangles,
and the six polyhedral angles are equal.
Why?
\end{minipage}%
% [** Illustration]
\hfill\raisebox{-0.125in}{\smash{\Input{061b}{1.5in}}}

\NewFolio %% -----File: 062.png---Folio 54-------

\Chapter{BOOK III}{Prisms and Cylinders}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{162.}{Closed Prismatic Surface.} Given a convex polygon and
\index{Prismatic@\textbf{Prismatic surface}}%{\Pnum{162}}%
\index{Surface@\textbf{Surface}!prismatic}%{\Pnum{162}}%
a straight line not in its plane. If the
line moves so as to remain parallel to
itself while it touches the boundary of
the polygon and is made to traverse it
completely, the line is said to generate
a \emph{closed prismatic surface}.

\Indent\Paragraph{163.}{Generator. Element.} The moving
\index{Generator@\textbf{Generator}!of a prismatic surface}%{\Pnum{163}}%
\index{Prismatic@\textbf{Prismatic surface}!generator of}%{\Pnum{163}}%
line is the \emph{generator} of the surface,
and the generator in any one of its positions
is an \emph{element} of the surface.
\end{minipage}%
% [** Illustration]
\hfill\Input{062a}{1.5in}

\Paragraph{164.}{Prism.} A polyhedron bounded by a prismatic surface
\index{Prism@\textbf{Prism}}%{\Pnum{164}}%
\index{Section@\textbf{Section}!of a prism}%{\Pnum{164}}%
and two parallel plane sections cutting all
its elements is called a \emph{prism}.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{165.}{Bases. Lateral Surface. Edges.}
\index{Base@\textbf{Base}!of a prism}%{\Pnum{165}}%
\index{Prism@\textbf{Prism}!bases of}%{\Pnum{165}}%
\index{Prism@\textbf{Prism}!lateral edges of}%{\Pnum{165}}%
\index{Lateral@\textbf{Lateral}!edges}%{\Pnum{165}}%
\index{Lateral@\textbf{Lateral}!faces}%{\Pnum{165}}%
\index{Lateral@\textbf{Lateral}!surfaces}%{\Pnum{165}}%
The two parallel cross-sections which
bound a prism are its \emph{bases} and the other
faces form its \emph{lateral surface}. The \emph{edges}
are the lines in which its lateral faces meet.

\Indent\Paragraph{166.}{Right Section. Right Prism.} A \emph{right section} of a prism
cuts all its edges at right angles. A \emph{right prism}
is one whose bases are right sections.
\index{Prism@\textbf{Prism}!right}%{\Pnum{166}}%
\index{Right@\textbf{Right}!prism}%{\Pnum{166}}%
\index{Right@\textbf{Right}!section}%{\Pnum{166}}%
\end{minipage}%
% [** Illustration]
\hfill\Input{062b}{1.25in}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1in}
\Indent\begin{Corollary}{167.}{1.}
The lateral faces of a
prism are parallelograms.
\end{Corollary}

\Indent\begin{Corollary}{168.}{2.}
The lateral edges of a
prism are equal and parallel.
\end{Corollary}

\Indent\begin{Corollary}{169.}{3.}
The lateral faces of a right prism are
rectangles.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\Input{062c}{1in}

\NewFolio %% -----File: 063.png---Folio 55-------

\Section{CLASSIFICATION OF PRISMS}
\SecToc{Properties of Prisms}

\Paragraph{170.}{Kinds of Prisms.} Prisms are classified according to
the form of their right sections, as \emph{triangular}, \emph{quadrangular},
\emph{pentagonal}, \emph{hexagonal}, etc. A \emph{regular prism} is one whose right
section is a regular polygon.
\index{Prism@\textbf{Prism}!hexagonal}%{\Pnum{170}}%
\index{Prism@\textbf{Prism}!quadrangular}%{\Pnum{170}}%
\index{Prism@\textbf{Prism}!regular}%{\Pnum{170}}%
\index{Prism@\textbf{Prism}!triangular}%{\Pnum{170}}%
\index{Regular@\textbf{Regular}!prisms}%{\Pnum{170}}%

\noindent\begin{minipage}[b]{\textwidth-3in}
\Indent\Paragraph{171.}{Altitude. Area.} The \emph{altitude} of a prism is the
\index{Altitude@\textbf{Altitude}!of a prism}%{\Pnum{171}}%
\index{Area@\textbf{Area}!of a prism}%{\Pnum{171}}%
\index{Faces@\textbf{Faces}!of a prism}%{\Pnum{171}}%
\index{Prism@\textbf{Prism}!altitude of}%{\Pnum{171}}%
\index{Prism@\textbf{Prism}!area of}%{\Pnum{171}}%
\index{Prism@\textbf{Prism}!lateral faces of}%{\Pnum{171}}%
perpendicular distance between
the planes of its
bases. The altitude of a
right prism is equal to its
edge.

\Indent The \emph{lateral area} of a
prism is the sum of the
areas of its lateral faces.
\end{minipage}%
% [** Illustration]
\hfill\Input{063abc}{3in}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent The \emph{total area} is the sum of its lateral area and the areas
of its bases.

\Indent\Paragraph{172.}{Truncated Prism.} A polyhedron which
\index{Polyhedrons@\textbf{Polyhedrons}!added}%{\Pnum{172}}%
\index{Prism@\textbf{Prism}!truncated}%{\Pnum{172}}%
\index{Truncated@\textbf{Truncated}!prism}%{\Pnum{172}}%
is a part of a prism cut off by a plane meeting
all the lateral edges, but not parallel to
the base, is called a \emph{truncated prism}.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{063d}{1in}}

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent Two polyhedrons are said to be \emph{added} when
they are placed so that a face of one coincides
wholly or in part with a face
of the other, but otherwise
each lies outside the other.

\Indent\Paragraph{173.}{Parallelopiped.} A parallelopiped
is a prism whose
bases, as well as lateral faces,
are parallelograms.
\index{Parallelopiped@\textbf{Parallelopiped}!rectangular}%{\Pnum{173}}%
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.5in}{\smash{\Input{063e}{2in}}}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent A \emph{rectangular parallelopiped} has its bases and
all its faces rectangles.

\Indent A \emph{cube} is a parallelopiped whose bases and
faces are all squares.
\index{Cube@\textbf{Cube}}%{\Pnum{173}}%
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{063f}{1in}}

\NewFolio %% -----File: 064.png---Folio 56-------

\Section{PARALLEL CROSS-SECTIONS OF A PRISM ARE EQUAL}

\begin{Theorem}{174.}{I.}
The cross-sections of a prism
made by parallel planes are equal polygons.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{064ab}{}
\end{center}

\Objective{Given a prism cut by two parallel planes forming the polygons
$ABCDE$ and $A'B'C'D'E'$.}
{To prove that $ABCDE = A'B'C'D'E'$.}

\textbf{Outline of Proof}: (1)~Show that $AB = A'B'$, $BC = B'C'$, etc.,
by proving that $ABB'A'$, $BCC'B'$, etc., are $\Pgrams$.

(2)~Show that $\angle ABC = \angle A'B'C'$, $\angle BCD = \angle B'C'D'$, etc.

(3)~Hence show that $ABCDE$ and~$A'B'C'D'E'$ can be made
to coincide.

\begin{Corollary}{175.}{}
Every cross-section of a prism parallel to
the base is equal to the base.
\end{Corollary}

\begin{Sightwork}

\item[1.] Can a parallelopiped be a \emph{right} prism without being a \emph{rectangular}
parallelopiped? Illustrate.

\item[2.] Show that in a rectangular parallelopiped each edge is perpendicular
to the other edges which meet it.

\item[3.] Show that any section of a prism made by a plane parallel to an edge
is a parallelogram.

\begin{Suggestion}
Use in order §§~\Sref{89}, \Sref{84}, and~\Sref{92}.
\end{Suggestion}

\item[4.] Is it possible to construct a prism such that there is no plane perpendicular
to its edges and cutting all of them unless some of them are
extended?

\end{Sightwork}

\NewFolio %% -----File: 065.png---Folio 57-------

\Section{THE LATERAL AREA OF A PRISM}

\begin{Theorem}{176.}{II.}
\index{Prism@\textbf{Prism}!area of}%{\Pnum{176}}%
The lateral area of a prism is
equal to the product of a lateral edge and the perimeter
of a right section.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{065ab}{}
\end{center}

\begin{Suggestion}
Show that the lateral edges are mutually equal and that
the area of each face is the product of a lateral edge and one side of
the right-section polygon.

Complete the proof.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1in}
\Indent\begin{Remark}
\textit{Note}. The form of statement in this theorem is the usual abbreviation
for the more \emph{precise} form:

\Indent The lateral area of a prism is equal to the product of the \emph{numerical
measures} of a lateral edge and the perimeter of  a right
section.

\Indent Similar abbreviations are used throughout this text.
\end{Remark}

\Indent\begin{Corollary}{177.}{}
The lateral area of a right
prism is equal to the product of its altitude and the
perimeter of its base.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\Input{065c}{1in}

\begin{Sightwork}

\item[1.] If the lateral edge of a prism is $8$~inches and the perimeter of its
right section $30$~inches, what is the lateral area?

\item[2.] Of all prisms having an altitude $6$~inches and the perimeter of a
right section $24$~inches, which one has the smallest lateral area?

\begin{Suggestion}
Which one has the shortest lateral edge?
\end{Suggestion}

\item[3.] Why is not the lateral area of an oblique prism determined by the
perimeter of the base and a lateral edge?
\end{Sightwork}

\NewFolio %% -----File: 066.png---Folio 58-------

\Section{CONDITIONS MAKING PRISMS EQUAL}

\begin{Theorem}{178.}{III.}
Two prisms are equal if three
faces having a common vertex in the one are equal
respectively to three faces having a common vertex in
the other, and similarly placed.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{066ab}{}
\end{center}

\Objective{Given the three faces meeting at~$B$ in prism~$P$ equal respectively
to the three faces meeting at~$B'$ in prism~$P'$, and similarly placed.}
{To prove that~$P$ can be made to coincide with $P'$.}

\textbf{Outline of Proof}: Trihedral $\Angles B$ and~$B'$ are equal. Why?

Now apply the two prisms, making $B$ coincide with $B'$, and
then show in detail that:

(1)~The lower bases coincide.

(2)~The lateral faces at $B$ and $B'$ coincide.

(3)~The upper bases coincide.

(4)~All the lateral faces coincide.

\noindent\begin{minipage}[b]{\textwidth-2.25in}
\Indent\begin{Corollary}{179.}{1.}
Two truncated prisms are equal under the
conditions of \upshape§~\Sref{178}.
\end{Corollary}
\end{minipage}%
% [** Illustrations]
\hfill\smash{\Input{066cd}{2in}}

\begin{Corollary}{180.}{2.}
Two right prisms are equal if they have
equal bases and equal altitudes.
\end{Corollary}

\NewFolio %% -----File: 067.png---Folio 59-------

\Section{OPPOSITE FACES OF A PARALLELOPIPED}

\begin{Theorem}{181.}{IV.}
Opposite faces of a parallelopiped
are equal and parallel.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{067a}{0.4\textwidth}
\end{center}

\begin{Suggestion}[s]
Consider the opposite faces $ABFE$ and $DCGH$.

(1)~Show that the sides of the angles $ABF$ and~$DCG$ are parallel, and
hence that the planes determined by them are parallel.

(2)~Show that these faces are equal.

In like manner argue about any other pair of opposite faces.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Corollary}{182.}{}
Any section of a
parallelopiped made by a plane cutting
four parallel edges is a parallelogram.
\end{Corollary}

\begin{Suggestion}
In the figure show that $RQ \parallel ST$
and $QT \parallel RS$.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfill\Input{067b}{1.25in}

\begin{Sightwork}

\item[1.] A section made by a plane passed through two diagonally opposite
edges of a parallelopiped is a parallelogram, for example, the section
through $DH$ and $BF$ in the figure of §~\Sref{181}.

\item[2.] Show that any two of the four diagonals of a parallelopiped bisect
each other.

\begin{Suggestion}
Make use of example~1.
\end{Suggestion}

\item[3.] If the parallelopiped in §~\Sref{182} is rectangular, is the section necessarily
a rectangle? May it be so?

\item[4.] Show that the section of a cube made by a plane through two diagonally
opposite edges is a rectangle. Why can it not be a square? Can
you construct a parallelopiped in which such a section would be a square?

\end{Sightwork}

\NewFolio %% -----File: 068.png---Folio 60-------

\Section{PLANE THROUGH OPPOSITE EDGES OF A PARALLELOPIPED}

\begin{Theorem}{183.}{V.}
A plane through diagonally
opposite edges of a parallelopiped divides it into two
triangular prisms having equal right sections and equal
lateral edges.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{068a}{0.4\textwidth}
\end{center}

\Objective{Given the parallelopiped $ABCD-F$ with opposite edges $AE$
and $CG$.}
{To prove that the plane through $AE$ and $CG$ divides it into
two triangular prisms having equal lateral edges and equal right
sections.}

\Proof: By §~\Sref{182} the right section $KLMO$ of the given prism
is a parallelogram. Hence $KM$ divides it into two equal
triangles. Hence the right sections of the prisms $ABC-F$ and
$ACD-H$ are equal.

The lateral edges $BF$ and $DH$ are equal by §~\Sref{168}.

\begin{Sightwork}

\item[1.] If a right section of a prismatic space is a parallelogram, what kind
of prism will be cut out by two right sections?

\item[2.] If a right section of a prismatic space is a rectangle, what kind of
prism will be cut out by two right sections?

\item[3.] If a right section of a prismatic space is a regular hexagon, what
kind of prism will be cut out by two right sections?

\item[4.] What kind of prism will be cut out by two right sections of any
closed prismatic surface?

\end{Sightwork}

\NewFolio %% -----File: 069.png---Folio 61-------

\begin{Exercises}

\item[1.] The face angles of one trihedral angle of a parallelopiped
are $65$°, $70$°, and~$75$° respectively. What are the face angles of
the other trihedral angles?

\item[2.] Show that two prisms cut from the same prismatic
space and having equal lateral edges have equal lateral areas.

\item[3.] Show that the diagonals of a rectangular
parallelopiped are all equal to each other.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[4.] Show that the square on the diagonal of
a rectangular parallelopiped is equal to the
sum of the squares on the sides meeting at a
vertex from which it is drawn.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{069a}{1.375in}}

\begin{Remark}
E.g.~in the figure $\overline{AG}^2 = \overline{AC}^2 + \overline{CG}^2 = \overline{AB}^2 + \overline{BC}^2 + \overline{CG}^2$.
\end{Remark}

\item[5.] Find the ratio of the diagonal to one edge of a cube.

\item[6.] Find the edge of a cube whose diagonal is $14$~inches.
Find the diagonal of a cube whose edge is $16$~inches.

\item[7.] Find a diagonal of a rectangular parallelopiped whose
edges are $6$, $8$, and~$10$ inches respectively.

\item[8.] Are the diagonals of a cube perpendicular to each other?

\item[9.] In a rectangular parallelopiped the diagonal of the base
is $12$~in.\ and the altitude is $5$~in. Find the diagonal of the
parallelopiped.

\item[10.] If two equal right prisms whose bases are equilateral
triangles are placed together so as to form one prism whose
base is a parallelogram, compare the lateral area of the prism
so formed with the sum of the lateral areas of the original
prisms.

\item[11.] A right prism whose bases are regular hexagons is
divided into six prisms whose bases are equilateral triangles.
Compare the lateral area of the original prism with the sum of
the lateral areas of the resulting prisms.

\end{Exercises}

\NewFolio %% -----File: 070.png---Folio 62-------

\Section{VOLUME OF A RECTANGULAR PARALLELOPIPED}

\Paragraph{184.}{} Thus far certain properties of prisms have been studied,
\index{Parallelopiped@\textbf{Parallelopiped}!volume of}%{\Pnum{184}}%
\index{Prism@\textbf{Prism}!volume of}%{\Pnum{184}}%
\index{Volume@\textbf{Volume}!of a prism}%{\Pnum{184}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{184}}%
but no attempt has been made to \emph{measure the space occupied} by
such a solid. For this purpose we consider first a rectangular
parallelopiped.

\Paragraph{185.}{Numerical Measure.} In case each edge of a rectangular
parallelopiped is commensurable with a unit segment, the number
of times which a unit cube is contained in it is the \emph{numerical
measure} or the \emph{volume} of the parallelopiped.

\Paragraph{186.}{The Commensurable Case.} In the commensurable
case just described, the volume
is easily computed.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.375in}
\Indent\begin{Remark}
E.g.~if in the figure one edge $AC$ is $4$~units, and
an adjoining edge $AB$ is $3$~units, then a cube as $AK$,
whose edge is one unit, may be laid off $4$~times along
$AC$ and a tier of $3\cdot 4 = 12$ such cubes will adjoin
the face~$AD$. Since the edge~$BE$ is $5$~units long, $5$
such tiers will exactly fill the space within the solid. That is, $5\cdot 3\cdot 4 = 60$
is the number of cubic units in the solid.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfill\Input{070a}{1.25in}

Hence in this case
\[
\textit{Volume} = \textit{Length}\/ × \textit{Width}\/ × \textit{Height}.
\]

\begin{Remark}
Again, if the given dimensions are $3.4$, $2.6$, $4.5$ decimeters respectively,
then unit cubes, with edge one \emph{decimeter}, cannot be made to fill exactly
the space inclosed by the figure, but cubes with edge each one \emph{centimeter}
will do so, giving $34$, $26$, and~$45$ respectively along the three edges of the
figure.

Therefore, the volume is

$34\cdot 26\cdot 45 = 39,780$ cubic centimeters, or $39.78$ cubic decimeters.
\end{Remark}

Hence in this case also
\[
\textit{Volume} = \textit{Length}\/ × \textit{Width}\/ × \textit{Height}.
\]

\Paragraph{187.}{Formula for Volume of a Rectangular Solid.} We therefore
conclude that the volume of a rectangular parallelopiped with
commensurable dimensions is given by the formula
\[
\textbf{Volume} = \textbf{Length} × \textbf{Width} × \textbf{Height}.
\]

\NewFolio %% -----File: 071.png---Folio 63-------

\Section{VOLUME OF A RECTANGULAR SOLID}

\Paragraph{188.}{The Incommensurable Case.} If a rectangular parallelopiped
\index{Incommensurables@\textbf{Incommensurables}}%{\Pnum{188}}%
\index{Prism@\textbf{Prism}!volume of}%{\Pnum{188}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{188}}%
is such that any two of its dimensions are incommensurable
with each other, then there is no unit cube, however
small, in terms of which the volume can be exactly expressed.

But by choosing a unit cube sufficiently small we may determine
the volume of a rectangular solid which differs from the
given one by as little as we please.

\begin{Remark}
E.g.,~if the length is $5$~inches and the width is $\sqrt{3}$~inches, then the
base cannot be exactly covered with equal cubes, however small.

But since $\sqrt{3} = 1.732\dots$, if we take as a unit of volume a cube one one-hundredth
of an inch on a side, then the base of a rectangular solid whose
length is $5$~inches and whose width is $1.73$~inches can be exactly covered
with a layer of these cubes and the number of cubes in such a layer is
\[
500 × 173 = 86,500.
\]

If the height is $2$~inches, then $200$~layers will just reach the top, making
\[
500 × 173 × 200 = 17,300,000~\text{cubes}.
\]

And since $1,000,000$ such small cubes make a cubic inch, the volume
of this solid is $17.3$~cubic inches.

The portion of the given solid not thus filled is $5$~inches long, $2$~inches
high and less than $.003$ of an inch in thickness.

Hence, its volume is less than $5\times 2\times .003 = .03$ of a cubic inch.

By expressing $\sqrt{3}$ to further places of decimals and then using smaller
and smaller units of volume, successive rectangular solids may be found
which differ less and less from the given solid.
\end{Remark}

The foregoing considerations constitute an informal proof of
the following theorem:

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Theorem}{189.}{VI.}
The volume
of a rectangular parallelopiped is
equal to the product of its length,
width, and height.
\end{Theorem}
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.125in}{\Input{071a}{1.5in}}}
\[
\text{\bfseries In symbols},\  V = l\cdot w\cdot h.
\]

\begin{Remark}
The above argument shows that this theorem holds for all rectangular
parallelopipeds used in the process of approximation, and hence it applies
to all practical measurements of the volumes of such solids.
\end{Remark}

\NewFolio %% -----File: 072.png---Folio 64-------

\begin{Corollary}{190.}{1.}
The  volume of a rectangular parallelopiped
is equal to the product of the numerical measures of its base
and altitude.
\end{Corollary}

\begin{Corollary}{191.}{2.}
If two rectangular parallelopipeds have
two dimensions respectively equal to each other, their volumes are
in the same ratio as their third dimensions; and if they have one
dimension the same in each, their volumes are in the same ratio
as the products of their other two dimensions.
\end{Corollary}

\begin{Remark}
For if $V$ and $V'$ are the volumes, and $a$, $b$, $c$, and $a'$, $b'$, $c'$ the dimensions,
then $\dfrac{V}{V'} = \dfrac{a\cdot b\cdot c} {a'\cdot b'\cdot c'} = \dfrac{a}{a'}$ if $b = b'$ and $c = c'$; or $\dfrac{V}{V'} = \dfrac{a\cdot b}{a'\cdot b'}$ if $c = c'$.
\end{Remark}
\medskip % [** PP: Manually-added space]

\Paragraph{192.}{Volumes of Prisms in General.} From the formula for
volumes of rectangular solids, $V = l\cdot w\cdot h$, we deduce the volumes
of prisms in general by means of the principle:

\textit{Two polyhedrons have the same volume if they can be made to
coincide, or if they can be divided into parts which can be made to
coincide in pairs}.
\index{Equal@\textbf{Equal}!volumes}%{\Pnum{192}}%
\index{Polyhedrons@\textbf{Polyhedrons}!equal, equivalent}%{\Pnum{192}}%
\index{Solids@\textbf{Solids}!equal}%{\Pnum{192}}%
\index{Solids@\textbf{Solids}!equivalent}%{\Pnum{192}}%
\index{Volume@\textbf{Volume}!of a prism}%{\Pnum{192}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{192}}%

\begin{Remark}
The sign ${}={}$ between two polyhedrons means that they are equal in
all respects; that is, can be made to coincide. The word \emph{equivalent} is
used to mean \emph{equal in volume}.
\index{Equivalent@\textbf{Equivalent solids}}%{\Pnum{192}}%
\end{Remark}

\begin{Exercises}

\Page{63}%
\item[1.] The dimensions in inches of a rectangular parallelopiped
are $5$, $10$, and~$2\sqrt{2}$. Taking the approximate value of~$\sqrt{2}$ as
$1.4142$, find the approximate volume. Show that this differs
from the exact volume by less than $.001$ of a cubic inch, it being
given that $\sqrt{2}$ lies between $1.41421$ and~$1.41422$.

\item[2.] If the dimensions in inches of a rectangular parallelopiped
are $5$, $4\sqrt{2}$, and~$5\sqrt{3}$, find how near an approximation
to the volume is possible by taking $\sqrt{2}$ between $1.414$ and~$1.4143$
and $\sqrt{3}$ between $1.732$ and~$1.7321$.

\item[3.] If the dimensions in inches of a rectangular parallelopiped
are $3\sqrt{2}$, $3\sqrt{3}$, $2\sqrt{5}$, find the values to be used for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$,
to obtain the volume within $.001$ of a cubic inch.
\end{Exercises}

\NewFolio %% -----File: 073.png---Folio 65-------

\Section{VOLUME OF AN OBLIQUE PRISM}

\begin{Theorem}{193.}{VII.}
\index{Prism@\textbf{Prism}!volume of}%{\Pnum{193}}%
The volume of an oblique
prism is equal to that of a right prism having for its
base a right section of the oblique prism and for its
altitude a lateral edge of the oblique prism.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{073ab}{}
\end{center}
\vspace*{-0.125in}

\Objective{Given the oblique prism $AD'$ with $FGHJK$ a right section and
$F'G'H'J'K'$ a right section of the prism extended so that $AA' = KK'$.}
{To prove that the oblique prism $AD'$ has the same volume as the
right prism $KH'$.}

\Proof: In the truncated prisms $AH$ and~$A'H'$,
base $ABCDE = {}$ base $A'B'C'D'E'$.\hfill §~\Sref{175}

In faces $ABFK$ and $A'B'F'K'$, we have
\begin{flalign*}
&&                 AB &= A'B',\quad \angle 1 = \angle 3,\ \angle 2 = \angle 4, && \tag*{Why?} \\
&\text{and}\quad & AK &= A'K' \text{ and } BF = B'F'. && \tag*{Why?}
\end{flalign*}

$\therefore ABFK$ and $A'B'F'K'$ will coincide and are equal.
\begin{flalign*}
&\indent\text{Likewise},& AKJE &= A'K'J'E'. && && \\
&&          \therefore\quad AH &= A'H'.     && && \tag*{§~\Sref{179}} \\
&\indent\text{But the given prism} & AD' &= AH + KD', && && \\
&\text{and the right prism} & KH' &= A'H' + KD'. && && \\
&&          \therefore\quad AD' &= KH'.     && && \tag*{\QED}
\end{flalign*}

\begin{Corollary}{194.}{}
Two prisms having equal lateral edges
and equal right sections are equal in volume.
\end{Corollary}

\NewFolio %% -----File: 074.png---Folio 66-------

\Section{VOLUME OF ANY PARALLELOPIPED}

\begin{Theorem}{195.}{VIII.}
The volume of any parallelopiped
is equal to the product of its base and altitude.
\end{Theorem}

% [** Illustration]
\vspace*{-0.25in}
\begin{center}
\Input{074a}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given any parallelopiped $P$ with area of base~$b$ and altitude~$h$.}
{To prove that its volume is equal to $b × h$.}

\Proof: Considering face $AK$ as the base of prism~$P$, produce
the four edges parallel to~$AB$, and lay off $A'B' = AB$.

Through $A'$ and $B'$ erect planes $\perp A'B'$, thus cutting off
the right prism~$Q$ with $A'L$ as one base.

Now considering $CL$ as the base of prism~$Q$, produce the
four edges $\parallel CB'$ and lay off $C'D = CB'$.

At $C'$ and $D$ erect planes $\perp C'D$, cutting off the right
prism~$R$, which is a rectangular parallelopiped with base~$b''$.

Now prove in detail: (1)~$h = h' = h''$,\quad (2)~$b = b' = b''$,
\begin{align*}
(3)\quad\text{Vol.\ } P &= \text{Vol.\ } Q = \text{Vol.\ } R,
\tag*{§~\Sref{193}} \\
(4)\quad\text{Vol.\ } R &= b'' h'',\qquad (5)\quad \text{Vol.\ } P = bh.
\end{align*}

\begin{Corollary}{196.}{}
Two parallelopipeds are equal in volume
if they have equal altitudes and bases of equal areas.
\end{Corollary}

\NewFolio %% -----File: 075.png---Folio 67-------

\Section{VOLUME OF A TRIANGULAR PRISM}

\begin{Theorem}{197.}{IX.}
The volume of a triangular
prism is equal to the product of its base and altitude.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{075a}{0.4\textwidth}
\end{center}

\Objective{Given the triangular prism whose base is~$ABC$.}
{To prove that the volume is equal to the area of $\triangle ABC$ multiplied
by the altitude~$h$.}

\Proof: Complete the $\Pgrams ABCD$ and $EFGH$ and draw~$DH$.
Now show that $CDHG$ and~$ADHE$ are $\Pgrams$, and hence that
$BH$ is a parallelopiped.\hfill \mbox{Use §~\Sref{43}~(2)}

Let $KLMO$ be a right section of~$BH$, and let~$KM$ be the
line in which the plane $ACGE$ cuts the plane $KLMO$.

Then (1)~prism $ABC-F = \tfrac{1}{2}$ prism~$BH$,
\hfill §§~\Sref{183},~\Sref{194}

(2)~But prism $BH = h × \text{area of } \Pgram ABCD$,
\hfill §~\Sref{196}

(3)~Hence, prism $ABC - F = h × \tfrac{1}{2} \text{area of } \Pgram ABCD$, that
is, prism $ABC - F = h × \text{area of } \Pgram ABC$.

Therefore the volume of prism $ABC-F$ is equal to the area
of its base times its altitude.
\QED

\begin{Sightwork}

\item[1.] Find the volume of a prism whose altitude is $8$~in.\ and whose base
is a right triangle with legs $5$~in.\ and $6$~in.\ respectively.

\item[2.] Find the volume of a right prism whose altitude is $6$~in.\ and whose
base is a rectangle with sides $3$~in.\ and $5$~in.
\end{Sightwork}

\NewFolio %% -----File: 076.png---Folio 68-------

\Section{VOLUME OF ANY PRISM}

\begin{Theorem}{198.}{X.}
\index{Prism@\textbf{Prism}!volume of}%{\Pnum{198}}%
The  volume  of any prism is
equal to the product of its base and altitude.
\end{Theorem}

% [**Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{076ab}{}
\end{center}
\vspace*{-0.25in}

\textbf{Outline of Proof}: Any prism, can be divided into triangular
prisms by planes passing through one edge and each of the
other non-adjacent edges.

The altitudes of the triangular prisms are the same as that
of the given prism, and the sum of their bases is equal to the
base of the given prism.

Now use §~\Sref{197} and complete the proof.

\begin{Corollary}{199.}{1.}
The product of the base and altitude of
any prism is equal to the product of a lateral edge and the area
of a right section.
\end{Corollary}

\begin{Remark}
For each equals the volume of the prism. See §§~\Sref{198},~\Sref{193}.
\end{Remark}

\begin{Corollary}{200.}{2.}
If two prisms have equivalent bases,
their volumes are in the same ratio as their altitudes; and if they
have equal altitudes, their volumes are in the same ratio as the
areas of their bases.
\end{Corollary}

\begin{Sightwork}

\item[1.] Find the volume of a prism whose altitude is $8$~in.\ and the area of
whose base is $42$~sq.~in.

\item[2.] The volume of a prism is $264$~cu.~in.\ and its altitude is $8$~in. Find
the area of its base.
\end{Sightwork}

\NewFolio %% -----File: 077.png---Folio 69-------

\begin{Exercises}

\item[1.] The theorem that the volume of any prism is equal to
the product of its base and altitude is of great importance.
What theorems of Book~III have been used directly or indirectly
in proving it?

\item[2.] What dimensions of a prism must be known in order to
determine its lateral area by means of the preceding theorems
of Book~III? What dimensions must be known to determine
its volume?

\item[3.] The edge of a cube is~$e$. Find the total surface and the
volume in terms of~$e$.

\item[4.] Find the edge of a cube if its total area is equal numerically
to its volume, an inch being used as the unit.

\item[5.] Find the volume of a regular right triangular prism
whose edges are all equal to $6$~inches.

\item[6.] Find the volume of a regular right hexagonal prism
whose edges are all equal to $10$~inches.

\item[7.] A side of the base of a regular right hexagonal prism is
$3$~inches. Find its altitude if its volume is $54\sqrt{3}$ cubic inches.
What is the total area of this prism?

\item[8.] The volume of a triangular prism is equal to the area of
one lateral face multiplied by half the perpendicular distance
of this face from the remaining edge. Prove.

\item[9.] The volume of a regular right prism is equal to the
lateral area multiplied by half the apothem of the base.
Prove.

\item[10.] Prove that the sum of the squares of the four diagonals
of a rectangular parallelopiped is equal to the sum of the
squares of the twelve edges of the parallelopiped.

\item[11.] A prismatic space is cut by two pairs of parallel planes
which are the same distance apart. Does it follow that the volumes
of the prisms thus formed are equal? The lateral areas?
\end{Exercises}

\NewFolio %% -----File: 078.png---Folio 70-------

\Section{CYLINDERS}
\SecToc{Properties of Cylinders}

\Paragraph{201.}{Curved Surface.} A surface no part of which is plane is
called a \emph{curved surface}.
\index{Curved surface@\textbf{Curved surface}}%{\Pnum{201}}%
\index{Surface@\textbf{Surface}!curved}%{\Pnum{201}}%

\begin{Remark}
E.g.~the surface of an eggshell or of a stovepipe is a curved surface.
\end{Remark}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{202.}{A Closed Plane Curve.} A curve which can be traced continuously
\index{Curves@\textbf{Curves}!convex, closed}%{\Pnum{202}}%
by a point moving in a
plane so as to return to its original
position without crossing its path
is a \emph{closed plane curve}.
\end{minipage}%
% [** Illustrations]
\hfill\smash{\Input{078a}{2in}}

A closed plane curve is \emph{convex}
if it can be cut by a straight line in not more than two
points.\footnoteTrans{See \Tref{078a}{Transcriber's Notes}.}
\Erratum{err078a}%
% [** F1: Modern parsing makes this mathematically incorrect]

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{203.}{A Cylindrical Surface.} If a straight
\index{Cylindrical@\textbf{Cylindrical surface}}%{\Pnum{203}}%
\index{Cylindrical@\textbf{Cylindrical surface}!element of}%{\Pnum{203}}%
\index{Cylindrical@\textbf{Cylindrical surface}!generator of}%{\Pnum{203}}%
\index{Generator@\textbf{Generator}!of a cylindrical surface}%{\Pnum{203}}%
\index{Surface@\textbf{Surface}!cylindrical}%{\Pnum{203}}%
line moves so as to remain parallel to
itself, while it always touches a closed
convex plane curve and is made to traverse
it completely, the line is said to
generate a \emph{closed convex cylindrical surface}.
The moving line is the \emph{generator}, and the
generator in any one of its positions is an
\emph{element} of the surface.

\Indent\Paragraph{204.}{Cylinder.} A solid bounded by a
\index{Cylinder@\textbf{Cylinder}!bases of}%{\Pnum{204}}%
\index{Cylinder@\textbf{Cylinder}!right}%{\Pnum{204}}%
\index{Section@\textbf{Section}!of a cylinder}%{\Pnum{204}}%
cylindrical surface and two parallel plane
sections cutting all its elements is called
a \emph{cylinder}.

\Indent\Paragraph{205.}{Bases. Lateral Surface.} The two
\index{Base@\textbf{Base}!of a cylinder}%{\Pnum{205}}%
\index{Cylinder@\textbf{Cylinder}!bases of}%{\Pnum{205}}%
parallel cross-sections which bound a cylinder
are its \emph{bases}, and the curved surface
is its \emph{lateral surface}.
\end{minipage}%
% [** Illustration]
\hfill\Input{078bc}{1.25in}

% [** Illustration]

\Paragraph{206.}{Element. Altitude.} That part of the generator of a
\index{Altitude@\textbf{Altitude}!of a cylinder}%{\Pnum{206}}%
\index{Cylinder@\textbf{Cylinder}!element of}%{\Pnum{206}}%
\index{Element@\textbf{Element}!of a cylinder}%{\Pnum{206}}%
cylindrical surface which is included between the bases of a
cylinder is called an \emph{element} of the cylinder. The \emph{altitude} of
a cylinder is the perpendicular distance between its bases.

\NewFolio %% -----File: 079.png---Folio 71-------

\Section{CIRCULAR CYLINDERS}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{207.}{Right Section.} A \emph{right section of a cylinder} is made by
\index{Right@\textbf{Right}!section}%{\Pnum{207}}%
a plane cutting each of its elements at right angles.

\Indent\Paragraph{208.}{Circular Cylinders.} A \emph{circular
cylinder} is one whose right section is
a circle.
\index{Circular@\textbf{Circular}!cylinder}%{\Pnum{208}}%
\index{Cylinder@\textbf{Cylinder}!circular}%{\Pnum{208}}%
\index{Radius@\textbf{Radius}!of a circular cylinder}%{\Pnum{208}}%
\index{Right@\textbf{Right}!cylinder}%{\Pnum{208}}%
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{079a}{1.5in}}

\Indent The radius of a circular cylinder is
the radius of its right section.

A cylinder whose elements are at
right angles to its bases is called a
\emph{right cylinder}. Otherwise it is an \emph{oblique cylinder}. A right
cylinder whose bases are circles is a \emph{right
circular cylinder}.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{209.}{Axis of a Cylinder.} The line passing
\index{Axis@\textbf{Axis}!of a cylinder}%{\Pnum{209}}%
through the centers of two right sections of
a circular cylinder is the \emph{axis} of the cylinder.

\Indent\begin{Remark}
A right circular cylinder may be generated by
revolving a rectangle about one of its sides as an
axis. The side opposite the axis generates the lateral surface, and the
sides adjacent to the axis generate the bases.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfill\Input{079b}{1in}

\begin{Sightwork}

\item[1.] What is the locus of all points in space which are at a perpendicular
distance of $6$~in.\ from a straight line $10$~in.\ long?
\index{Loci@\textbf{Loci problems}}%

\item[2.] Show that any two elements of a cylinder determine a plane section
of the cylinder which cuts the two bases in parallel lines.

\item[3.] If every plane determined by two elements of a cylinder is perpendicular
to its bases, what kind of cylinder is it?

\item[4.] If one right section of a cylinder is a circle, what would appear to
be true of all its right sections? For proof see §~\Sref{213}.

\item[5.] If a circular cylinder is oblique, are its bases circles?

\item[6.] If an oblique cylinder has a circular base, is its right section a
circle?
\end{Sightwork}

\NewFolio %% -----File: 080.png---Folio 72-------

\Section{LONGITUDINAL SECTIONS OF A CYLINDER}

\begin{Theorem}{210.}{XI.}
If a plane contains an element
of a cylinder and meets it in any other point, then it
contains another element also, and the section is a
parallelogram.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{080ab}{}
\end{center}

\Objective{Given the cylinder $AC$ and a plane~$M$ containing the element~$AD$
and another point as $P$ or~$P'$.}
{To prove that it contains another element and that the section
containing these two elements is a parallelogram.}

\Proof: \Case{(1)}{When the point~$P$ is in the lateral surface.}

Let $BC$ be the element through~$P$.

Then $BC\parallel AD$ and they determine a plane~$M$.\hfill §§~\Sref{203},~\Sref{73}

Also $AB\parallel DC$ and $ABCD$ is a parallelogram.\hfill §~\Sref{92}

\Case{(2)}{When the point $P'$ is in one of the bases.}

Draw $AB$ through $P'$. Then by~(1) the plane~$M$ contains
the element~$BC$ and the section is a
parallelogram.
\QED

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{211.}{Plane Tangent to a Cylinder.} If a
\index{Tangent@\textbf{Tangent}!to a cylinder}%{\Pnum{211}}%
plane contains an element of a cylinder
and no other point in it, the plane is said
to be \emph{tangent to the cylinder}, and the element
is called the \emph{line of contact}.

\Indent It may be noted that a cross-section of a cylinder and its
tangent plane consists of a plane curve and a line tangent to it.
\end{minipage}%
% [** Illustration]
\hfill\Input{080c}{1.25in}

\NewFolio %% -----File: 081.png---Folio 73-------

\Section{THE BASES OF A CYLINDER ARE EQUAL}

\begin{Theorem}{212.}{XII.}
The bases of a cylinder are
equal plane figures.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{081ab}{0.6\textwidth}
\end{center}

\Objective{Given a cylinder with the bases $b$ and $b'$.}
{To prove that $b = b'$.}

\Proof: Take any three points $A$, $B$, $C$ in the rim of the base~$b$
and draw elements through these points, meeting the base~$b'$
in $D$, $E$, $F$.

Show that $\triangle ABC = \triangle DEF$.\hfill Use~§§~\Sref{210},~\Sref{27}

Now, while the elements $AD$ and $BE$ remain fixed, conceive
$CF$ to generate the surface of the cylinder.

Evidently $\triangle ABC = \triangle DEF$ in every position of $CF$.

Hence, if base~$b'$ is applied to base~$b$ with these triangles
coinciding in \emph{one} position, they will coincide in every position
corresponding to the moving generator.

That is, $b'$ coincides with~$b$.
\QED

\begin{Remark}
The proposition of §~\Sref{212} may also be stated in the form of the following
corollary:
\end{Remark}

\begin{Corollary}{213.}{1.}
Parallel plane sections of a cylinder are
equal, if they cut all the elements.
\end{Corollary}

\begin{Corollary}{214.}{2.}
The axis of a circular cylinder passes
through the center of all its right sections.
\end{Corollary}

\NewFolio %% -----File: 082.png---Folio 74-------

\Section{MEASURING THE SURFACE AND VOLUME OF A CYLINDER}

\Paragraph{215.}{Area of a Curved Surface.} Thus far in geometry the
\index{Area@\textbf{Area}!of a curved surface}%{\Pnum{215}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{215}}%
word \emph{area} has been used in connection with plane figures only.
In some cases the computation of an area has been made by
an approximation process only, as in the case of some rectangles
and of the circle. Indeed, in these cases approximate measurement
only is possible, since no square unit exists in terms
of which such areas can be \emph{exactly} measured.

In the case of \emph{any curved surface} it is evident that approximate
measurement is the \emph{only kind possible} in terms of a plane
area unit, since no such unit, however small, can be made to
coincide with a part of such a surface.\footnoteTrans%
  {See \Tref{082a}{Transcriber's Notes}.}
\Erratum{err082a} % [** PP: Subtle geometric issue]

\Paragraph{216.}{Volume of a Solid Having a Curved Surface.} Since the
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{216}}%
unit of measure for solids is the cube, it is evident that a solid
having a curved surface cannot contain such a unit an integral
number of times. Hence approximate measurement only of
the volumes of such solids is possible.

\begin{Remark}
To measure approximately the area and volume of a cylinder use is
made of inscribed and circumscribed prisms.
\end{Remark}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{217.}{Inscribed Prisms.} A prism is said to be \emph{inscribed in a
cylinder} if its lateral edges are elements of the
cylinder, and their bases lie in the same
planes.
\index{Circumscribed@\textbf{Circumscribed}!cylinder}%{\Pnum{217}}%
\index{Cylinder@\textbf{Cylinder}!circumscribed}%{\Pnum{217}}%
\index{Inscribed@\textbf{Inscribed}!prism}%{\Pnum{217}}%
\index{Prism@\textbf{Prism}!inscribed}%{\Pnum{217}}%

\Indent\Paragraph{218.}{Circumscribed Prisms.} A prism is said
\index{Circumscribed@\textbf{Circumscribed}!prism}%{\Pnum{218}}%
\index{Cylinder@\textbf{Cylinder}!inscribed}%{\Pnum{218}}%
\index{Inscribed@\textbf{Inscribed}!cylinder}%{\Pnum{218}}%
\index{Prism@\textbf{Prism}!circumscribed}%{\Pnum{218}}%
to be \emph{circumscribed about a cylinder} if its
lateral faces are all tangent to the cylinder,
and their bases lie in the same planes.
\end{minipage}%
% [** Illustration]
\hfill\Input{082a}{1.25in}

\begin{Remark}
It is evident that by increasing the number of lateral faces of the inscribed
or circumscribed prisms the surface of the latter may be made to lie
as near the surface of the cylinder as we please. The lateral edges of the
prisms will remain equal to an element of the cylinder, while the right
sections of the prisms can be made to differ as little as we please from the
right section of the cylinder.
\end{Remark}

\NewFolio %% -----File: 083.png---Folio 75-------

\Section{LATERAL AREA OF A CYLINDER}

\Paragraph{219.}{Fundamental Assumption on the Area and Volume of a
Cylinder.} We now assume that
\index{Area@\textbf{Area}!of a cylinder}%{\Pnum{219}}%
\index{Axioms@\textbf{Axioms}}%{\Pnum{219}}%
\index{Cylinder@\textbf{Cylinder}!lateral surface of}%{\Pnum{219}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{219}}%

\textit{A cylinder has a definite lateral area and a definite volume
which may be approximated as nearly as we please by taking the
lateral areas and the volumes of successive inscribed or circumscribed
prisms}.

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{083ab}{}
\end{center}
\vspace*{-0.25in}

\Paragraph{220.}{Lateral Area of a Cylinder.} Since \textit{the lateral area of a
prism is equal to the product of a lateral edge and the perimeter
of a right section} (§~\Sref{176}), it follows that this theorem holds for
every inscribed or circumscribed prism used in approximating
the lateral area of a cylinder.

The foregoing considerations constitute an informal proof of
the following theorem:

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Theorem}{221.}{XIII.}
The lateral area
of a cylinder is equal to the product of an
element of the cylinder and the perimeter
of a right section.
\end{Theorem}
\end{minipage}%
% [** Illustration] % \smash *after* \raise
\hfill\smash{\raisebox{-0.375in}{\Input{083c}{1.5in}}}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\begin{Corollary}{222.}{1.}
If $r$ is the radius of a
right section of a circular cylinder, and~$e$ is an element,
the lateral surface is $S = 2 \pi re$ \Inum{(§~\Sref{176})}.
\end{Corollary}

\begin{Corollary}{223.}{2.}
If~$r$ is the radius and~$h$ is the
altitude of a right circular cylinder, then the lateral
surface is $S = 2 \pi rh$ \Inum{(§~\Sref{177})}.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.25in}{\Input{083d}{1in}}}

\NewFolio %% -----File: 084.png---Folio 76-------

\Section{VOLUME OF A CYLINDER}

\Paragraph{224.}{} Since \textit{the volume of a prism is equal to the product of its
altitude and the area of its base}, it follows that this theorem
holds for every inscribed or circumscribed prism used in approximating
the volume of a cylinder.
\index{Cylinder@\textbf{Cylinder}!volume of}%{\Pnum{224}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{224}}%

\begin{Remark}
\emph{All practical computations} of the surfaces and the volumes of cylinders
are \emph{approximations} based upon the propositions stated in §§~\Sref{220},~\Sref{224}.
\end{Remark}

The foregoing constitutes an informal proof of the following
theorem:

\begin{Theorem}{225.}{XIV.}
\index{Volume@\textbf{Volume}!of a cylinder}%{\Pnum{272}}%
The volume of a cylinder is
equal to the product of its altitude and the area of its
base.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1in}
\Indent\begin{Corollary}{226.}{1.}
If $r_1$ is the radius of the right
section of a circular cylinder and $e$ is an element, then
the volume is \Inum{(§~\Sref{199})} $V = \pi r_1^2 e$.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.25in}{\Input{084a}{1in}}}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Corollary}{227.}{2.}
If a cylinder of altitude~$h$
has a circular base whose radius is~$r_2$, then the volume
is
\[
V = \pi r_2^2 h.
\]
\end{Corollary}
\end{minipage}%
% [** Illustration]
\smash{\raisebox{0.125in}{\Input{084b}{1in}}}

\noindent\begin{minipage}[b]{\textwidth-1in}
\Indent\begin{Corollary}{228.}{3.}
If $h$ is the altitude and $r$ the
radius of the base of a right circular cylinder, then
the volume is $V = \pi r^2h$.
\end{Corollary}

\begin{Remark}
In this case $r = r_1 = r_2$, and $e = h$, and the formulas of
corollaries~1 and~2 become identical.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{084c}{1in}}

{\small%
\textit{Note}. The theorems of §§~\Sref{221}, \Sref{225} were stated (and are true) for any
cylinders whatever. However, the theorem of §~\Sref{225} is available for the
computation of the volume of a cylinder only in case the area of the base
or of the right section can be computed. And this is possible by
\emph{elementary methods} only in case these are circles.

The lateral area of a cylinder can be computed only in case the perimeter
of a right section can be found, and this is possible by elementary
methods only in case the right section is a circle.}

\NewFolio %% -----File: 085.png---Folio 77-------

\begin{Sightwork}

\item[1.] If a prism is inscribed in a cylinder, is its lateral area greater or
less than that of the cylinder? Is its volume greater or less than that of
the cylinder?

\item[2.] If a prism of say $1000$~faces is inscribed in a cylinder, would the
theorems of §~\Sref{221} and §~\Sref{225} apply directly to this prism? Would it be
easy to detect experimentally the difference between this prism and the
cylinder?

\item[3.] How would the right section of the prism and the cylinder of example~2
be related?

\item[4.] Discuss the questions in examples 1, 2, 3 as related to prisms circumscribed
about a cylinder.

\item[5.] What is the lateral area of a cylinder whose element is $8$~inches
and the perimeter of whose right section is $24$~inches?

\item[6.] What is the volume of a cylinder whose altitude is $10$~inches and
the area of whose base is $50$~square inches?

\item[7.] What is the volume of a cylinder whose edge is $4$~inches, and the
area of whose right section is $24$~square inches?

\end{Sightwork}

\begin{Exercises}

\item[1.] Find the volume of a right circular cylinder with radius~$5$
and altitude~$8$.

\item[2.] Find the volume of a circular cylinder if the radius of
a right section is~$6$ and the length of an element is~$10$.

\item[3.] Find the lateral area of a cylinder if the perimeter of a
right section is~$39$ and the length of an element is~$8$.

\item[4.] Find the lateral area of a right cylinder if the perimeter
of the base is~$30$ and the altitude is~$5$.

\item[5.] Find the total surface area of a right circular cylinder of
radius~$3$ and altitude~$5$.

\item[6.] Find the volume of an oblique cylinder whose base is a
circle of radius~$4$ and whose altitude is~$8$.
\end{Exercises}

\NewFolio %% -----File: 086.png---Folio 78-------


\Section{SUMMARY OF BOOK III}
\index{Summaries@\textbf{Summaries}}

\begin{Itemize}

\item[1.] Define prismatic surface and prismatic space.

\item[2.] Define prism, base of prism, lateral surface, right section,
right prism, parallelopiped, truncated prism.

\item[3.] Give the rule for finding the lateral area of an oblique
prism. How may this rule be modified in the case of a right
prism?

\item[4.] What can be said about opposite faces of a parallelopiped
as to shape and size?

\item[5.] Give two rules for finding the volume of an oblique
prism. See §§~\Sref{193}, \Sref{198}.

\item[6.] In proving the theorem on the volume of a prism, we
considered triangular prisms, general parallelopipeds, rectangular
parallelopipeds, and general prisms. In what order were
these taken? Why?

\item[7.] Show how you would find the approximate volume of a
rectangular parallelopiped whose dimensions are $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$.
Find this volume accurately to two places of decimals. Discuss
the problem of finding this volume correct to four places of
decimals. See exercises on page~\Pgref{63}.

\item[8.] Define closed plane curve, convex plane curve, cylindrical
surface, cylindrical space, cylinder, right cylinder, circular
cylinder, tangent plane to a cylinder.

\item[9.] Give the rule for finding the lateral surface of any
cylinder, of a right circular cylinder.

\item[10.] Give two rules for finding the volume of an oblique
cylinder. Show that these both hold true for a right circular
cylinder.

\item[11.] Describe what is meant by the statement ``the area and
the volume of a cylinder may be approximated by taking the
area and the volume of inscribed or circumscribed prisms.''
\end{Itemize}

\NewFolio %% -----File: 087.png---Folio 79-------

\Section{MISCELLANEOUS EXERCISES ON BOOK III}

\begin{Itemize}
\index{Problems@\textbf{Problems and Applications}|(}

\item[1.] If the lateral surface of a cylinder and the length of an
element are known, can the perimeter of a right section be
found? If the lateral area is~$400\pi$, and an element~$15$, find
the perimeter of a right section.

\item[2.] The diameter of a right circular cylinder is~$8$, and the
diagonal of the largest rectangle made by a vertical section is~$16$.
Find its altitude.

\item[3.] The volume of a right circular cylinder is $128\pi$~cubic
inches and its altitude is equal to its diameter. Find the
altitude and the diameter.

\item[4.] A rectangle whose sides are $a$ and $b$ is turned about the
side~$a$ as an axis and then about the side~$b$. Find the ratio of
the volumes of the two cylinders thus developed. Also find
the ratio of the total surfaces of these cylinders.

\item[5.] Find the diameter of a right circular cylinder if its
lateral area is equal numerically to its volume. Does the
result depend upon the altitude of the cylinder?

\item[6.] If the altitude of a right circular cylinder is equal to its
diameter, find the ratio of the numerical values of its total area
and its volume. Does this depend on the radius?

\item[7.] A regular hexagonal prism is inscribed in a right circular
cylinder whose altitude is equal to the diameter. Find the
difference between the volumes of the cylinder and the prism,
if a side of the hexagon is $4$~inches.

\item[8.] A cylindrical tank $8$~feet in diameter, partly filled with
water, is lying on its side. If the greatest depth of the water
is $6$~feet, what fraction of the volume
of the tank is filled with water?

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\item[9.] In the preceding problem find
the fraction of the volume occupied by
water if the width of the top of the
water along a right cross-section of the tank is $4$~feet.
\end{minipage}%
% [** Illustration]
\hfill\Input{087a}{1.75in}
\index{Problems@\textbf{Problems and Applications}|)}
\end{Itemize}

\NewFolio %% -----File: 088.png---Folio 80-------

\Chapter{BOOK IV}{Pyramids and Cones}

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{229.}{Pyramidal Surface.} Given a convex
\index{Generator@\textbf{Generator}!of a pyramidal surface}%{\Pnum{229}}%
\index{Pyramidal@\textbf{Pyramidal surface}}%{\Pnum{229}}%
\index{Pyramidal@\textbf{Pyramidal surface}!element of}%{\Pnum{229}}%
\index{Pyramidal@\textbf{Pyramidal surface}!vertex of}%{\Pnum{229}}%
\index{Surface@\textbf{Surface}!pyramidal}%{\Pnum{229}}%
\index{Vertex@\textbf{Vertex}!of a pyramid}%{\Pnum{229}}%
polygon and a fixed point~$P$ not in
its plane. If a line through the fixed
point moves so as to touch the boundary
of the polygon and is made to traverse
it completely, the line is said to generate
a convex \emph{pyramidal surface}.

\Indent The moving line is the \emph{generator} of the
surface, and in any of its positions it is
an \emph{element} of the surface. The fixed
point is the \emph{vertex} of the pyramidal
surface.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{088a}{1.75in}}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{230.}{Nappes.} A pyramidal surface has
two parts, one on each side of the vertex,
which are called \emph{nappes}.
\index{Pyramidal@\textbf{Pyramidal surface}!nappes of}%{\Pnum{230}}%
\index{Nappes@\textbf{Nappes}!of a pyramidal surface}%{\Pnum{230}}%

\Indent\Paragraph{231.}{Pyramid.} The solid bounded by a
\index{Pyramid@\textbf{Pyramid}}%{\Pnum{231}}%
\index{Section@\textbf{Section}!of a pyramid}%{\Pnum{253}}%
pyramidal surface and a plane section cutting
all the elements, and not passing
through the vertex, is called a \emph{pyramid}.
\end{minipage}%
% [** Illustration]
\hfill\Input{088b}{1.25in}

\begin{Sightwork}

\item[] If a plane cuts both nappes of a pyramidal surface, is it possible that it
should cut every element of the surface?

\begin{Suggestion}
\Page{79}%
If a plane~$M$ cuts both nappes, then a plane through the
vertex $\parallel M$ meets the surface in two elements which are $\parallel M$ and hence
are not cut by~$M$.
\end{Suggestion}
\end{Sightwork}

\NewFolio %% -----File: 089.png---Folio 81-------

\Section{CLASSIFICATION OF PYRAMIDS}
\SecToc{Properties of Pyramids}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{232.}{Faces. Edges.} The lateral surface of a pyramid
\index{Base@\textbf{Base}!of a pyramid}%{\Pnum{232}}%
\index{Faces@\textbf{Faces}!of a pyramid}%{\Pnum{232}}%
\index{Pyramid@\textbf{Pyramid}!base of}%{\Pnum{232}}%
\index{Pyramid@\textbf{Pyramid}!lateral faces of}%{\Pnum{232}}%
\index{Pyramid@\textbf{Pyramid}!triangular}%{\Pnum{232}}%
\index{Tetrahedron@\textbf{Tetrahedron}}%{\Pnum{232}}%
\index{Lateral@\textbf{Lateral}!edges}%{\Pnum{232}}%
\index{Lateral@\textbf{Lateral}!faces}%{\Pnum{232}}%
\index{Lateral@\textbf{Lateral}!surfaces}%{\Pnum{232}}%
is composed of \emph{triangular faces} having a
common vertex at the vertex of the pyramid, and
having as bases the sides of the polygon forming
the base. The sides common to two such triangles
are the \emph{edges} of the pyramid.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{089a}{1in}}

\textbf{Pyramids are classified}, according to the shape of the base, as
\emph{triangular}, \emph{quadrangular}, \emph{pentagonal}, etc.

\Paragraph{233.}{Tetrahedron.} A pyramid having a triangular base has
\index{Pyramid@\textbf{Pyramid}!altitude of}%{\Pnum{233}}%
in all four faces, and is called a \emph{tetrahedron}. In this case
every face is a triangle, and any one of them may be taken as
the base.

% [** PP: Changed bold to emph]
The \emph{altitude of a pyramid} is the perpendicular distance from
\index{Altitude@\textbf{Altitude}!of a pyramid}%{\Pnum{233}}%
the vertex to the plane of the base.

\Paragraph{234.}{Regular Pyramid.} A \emph{regular pyramid} is one whose base
\index{Pyramid@\textbf{Pyramid}!regular}%{\Pnum{234}}%
\index{Regular@\textbf{Regular}!pyramids}%{\Pnum{234}}%
\index{Right@\textbf{Right}!pyramid}%{\Pnum{234}}%
is a regular polygon such that the perpendicular from the vertex
upon it meets it at the center.

\Paragraph{235.}{Properties of a Regular Pyramid.}

(1)~\textit{The edges are equal to each other}.

\begin{Remark}
For they cut off equal distances from the foot of the perpendicular.
\end{Remark}

(2)~\textit{The faces are equal isosceles triangles}.

(3)~\textit{The altitudes of the triangular faces are equal to each other}.

\Paragraph{236.}{Slant Height.} The altitude of any one of the triangular
faces of a regular pyramid is called its \emph{slant height}.
\index{Slant@\textbf{Slant height}!of a pyramid}%{\Pnum{236}}%

\begin{Sightwork}

\item[1.] If the slant height of a regular pyramid and the apothem of its base
are given, how may its altitude be computed? If the altitude and
apothem are given, how may the slant height be found?

\item[2.] Can the expression ``slant height'' be applied to a pyramid if the
altitudes of its faces are not all equal? Discuss fully.
\end{Sightwork}

\NewFolio %% -----File: 090.png---Folio 82-------

\Section{LATERAL AREA OF A REGULAR PYRAMID}

\begin{Theorem}{237.}{I.}
\index{Area@\textbf{Area}!of a pyramid}%{\Pnum{237}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{237}}%
The lateral area of a regular
pyramid is equal to one half the product of its slant
height and the perimeter of the base.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{090ab}{}
\end{center}

\textbf{Suggestion for Proof}: Let $l = PK$ represent the slant height, $S$~the
lateral area, and $p$ the perimeter of the base. We are to show that
\[
S = \tfrac{1}{2} l\cdot p.
\]

Give all the steps.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\Paragraph{238.}{Pyramids Cut by a Plane.} The part
\index{Pyramid@\textbf{Pyramid}!frustum of}%{\Pnum{238}}%
\index{Pyramid@\textbf{Pyramid}!truncated}%{\Pnum{238}}%
\index{Truncated@\textbf{Truncated}!pyramid}%{\Pnum{238}}%
of a pyramid between its base and a plane
cutting all its edges is called a \emph{frustum} if
the cutting plane is parallel to the base,
and a \emph{truncated pyramid} if the cutting
plane is not parallel to the base.
%[** Illustration]

\Paragraph{239.}{The Parts of a Frustum.} The two
\index{Altitude@\textbf{Altitude}!of a frustum}%{\Pnum{239}}%
\index{Frustum@\textbf{Frustum}!of a pyramid}%{\Pnum{239}}%
\index{Slant@\textbf{Slant height}!of a frustum}%{\Pnum{239}}%
parallel faces of a frustum are its \emph{upper
and lower bases}, the perpendicular distance
between the bases is its \emph{altitude}, the trapezoidal
faces are its \emph{lateral surface}, and the
common altitude of these faces is the \emph{slant
height} of the frustum.
\end{minipage}%
% [** Illustration]
\hfill\Input{090cd}{1.25in}

\begin{Corollary}{240.}{}
The lateral area of the frustum of a regular
pyramid is equal to half the sum of the perimeters of the bases
multiplied by the slant height.
\end{Corollary}

\begin{Suggestion}
Use §~\Sref{62} and the figure in §~\Sref{238}.
\end{Suggestion}

\NewFolio %% -----File: 091.png---Folio 83-------

\Section{SECTION PARALLEL TO THE BASE OF A PYRAMID}

\begin{Theorem}{241.}{II.}
If a pyramid is cut by a plane
parallel to the base, then

\Inum{(1)}~the edges and altitude are divided proportionally;

\Inum{(2)}~the polygonal section is similar to the base;

\Inum{(3)}~the areas of this section and of the base are in the
same ratio as the squares of their perpendicular distances
from the vertex.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{091ab}{}
\end{center}

\textbf{Outline of Proof}: Let $A'B'C'D'E'$ be a section $\parallel ABCDE$.

(1)~To prove that $\dfrac{PK'}{PK} = \dfrac{PA'}{PA} = \dfrac{PB'}{PB}$, etc., pass another plane
through $P$ parallel to the base and then use §~\Sref{103}.

(2)~To prove that $ABCDE\sim A'B'C'D'E'$, we show that
$\angle A = \angle A'$, $\angle B = \angle B'$, etc., and also $\dfrac{A'B'}{AB} = \dfrac{B'C'}{BC}$, etc.

(3)~Calling the area of the section $b'$ and that of the base $b$, we
are to prove that $\dfrac{b'}{b} = \dfrac{\overline{PK'}^2}{\overline{PK}^2}$, and for this we need to show that
\[
\dfrac{A'B'}{AB} = \dfrac{PA'}{PA} = \dfrac{PK'}{PK}.
\]

Give all the steps in detail.

\NewFolio %% -----File: 092.png---Folio 84-------

\begin{Corollary}{242.}{}
If two pyramids have equal altitudes and
bases of equal areas lying in the same plane, the sections made
by a plane parallel to the plane of the bases have equal areas.
\end{Corollary}

% [** Illustration]
\begin{center}
\Input{092ab}{}
\end{center}

\begin{Suggestion}
If $t$ and $t'$ are the areas of the sections and $b$ and $b'$
those of the bases, show that $\dfrac{t}{b} = \dfrac{\overline{QL'}^2}{\overline{QL}^2}$, $\dfrac{t'}{b'} = \dfrac{\overline{PK'}^2}{\overline{PK}^2}$, and hence $\dfrac{t}{b} = \dfrac{t'}{b'}$, from
which it follows that $t = t'$ if $b = b'$.
\end{Suggestion}

\begin{Sightwork}

\item[1.] Find the lateral area of a regular pyramid if its slant height is $10$~in.\
and the perimeter of its base is $25$~in.

\item[2.] What is the perimeter of the base of a regular pyramid if its lateral
area is $72$~sq.~in.\ and its slant height is $8$~in.?

\item[3.] What is the slant height of a regular pyramid if its lateral area is
$160$~sq.~in.\ and the perimeter of its base is $20$~in.?

\item[4.] What is the perimeter of the base of a regular pyramid if its lateral
area is $250$~sq.~in.\ and its slant height is $20$~in.?

\item[5.] The area of the base of a pyramid is $36$~sq.~in.\ and its altitude is
$8$~in. What is the area of a section of this pyramid made by a plane
parallel to the base and $4$~in.\ from the vertex?

\item[6.] The area of the base of a pyramid is $64$~sq.~in.\ and its altitude is
$8$~in. Find the distance from the vertex to a plane parallel to the base if
the area of its section is $16$~sq.~in.

\item[7.] If the area of one section of a pyramid is double that of another,
both sections being parallel to the base, find the ratio of their distances
from the vertex. Find this ratio if the area of one section is three times
that of the other; also if it is four times.
\end{Sightwork}

\NewFolio %% -----File: 093.png---Folio 85-------

\Section{INSCRIBED AND CIRCUMSCRIBED PRISMS}

\Paragraph{243.}{Construction.} A triangular pyramid is cut by a series
\index{Circumscribed@\textbf{Circumscribed}!prism}%{\Pnum{243}}%
\index{Inscribed@\textbf{Inscribed}!prism}%{\Pnum{243}}%
\index{Prism@\textbf{Prism}!circumscribed}%{\Pnum{243}}%
\index{Prism@\textbf{Prism}!inscribed}%{\Pnum{243}}%
\begin{wrapfigure}[10]{r}{2.5in}
% [** Illustrations]
\Input{093ab}{2.5in}
\end{wrapfigure}
of planes parallel to the base, including one through the vertex
and also the one in
which the base lies.

Through the lines of
intersection of these
planes with one of the
faces, planes are constructed
parallel to the
opposite edge, thus forming
a set of prisms all
lying within the pyramid,
as $a'$, $b'$, $c'$ in pyramid
$P'$, or a set lying partly outside the pyramid, as $a$, $b$, $c$, $d$
in pyramid~$P$.

The inner prisms thus constructed are called \emph{a set of inscribed
prisms}, and the outer prisms are called \emph{a set of circumscribed
prisms}. For our purpose equidistant planes are used.

\begin{Remark}
This process may be repeated by doubling the number of planes drawn
parallel to the base and thus doubling the number of inscribed or circumscribed
prisms. By continuing in this way either set of prisms may
be made to coincide as nearly as we please with the pyramid.
\end{Remark}

\Paragraph{244.}{Fundamental Assumption on the Volume of a Pyramid.}
We now assume that
\index{Axioms@\textbf{Axioms}}%{\Pnum{244}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{244}}%

\textit{A pyramid has a definite volume which is less than the combined
volume of any set of circumscribed prisms and greater
than that of any set of inscribed prisms}.

\begin{Sightwork}

\item[1.] Show that in the above figure prisms $b$ and $a'$ are equal in volume
if $PM = P'M'$ and $\triangles ABC$ and $A'B'C'$ have equal areas. See §§~\Sref{188},
\Sref{242}. Also prove that volume $c = {}$ volume $b'$, etc.

\item[2.] Show that the sum of the volumes of the circumscribed prisms exceeds
the sum of the volumes of the inscribed prisms by the volume of~$a$.
\end{Sightwork}

\NewFolio %% -----File: 094.png---Folio 86-------

\Section{PYRAMIDS HAVING EQUAL VOLUMES}

\begin{Theorem}{245.}{III.}
If two triangular pyramids
have equal altitudes and bases of equal areas, their
volumes are equal.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{094ab}{}
\end{center}

\Objective{Given the pyramids $P$ and $P'$ in which altitudes $PM$ and $P'M'$
are equal, and the bases $ABC$ and $A'B'C'$ have equal areas.}
{To prove that $P$ and $P'$ have equal volumes.}

\Proof: If $P$ differs at all in volume from $P'$, let $P$ be the
greater, and let the difference be some fixed number $K$, so
that\hfill $\text{Vol.\ } P - \text{Vol.\ } P' = K$.\hfill    (1)

Construct a set of inscribed and circumscribed prisms as in
§~\Sref{243}.

Then\hfill $a' = b$, $b' = c$, $c' = d$.\hfill    Why?

Denote\hfill $a + b + c + d$ by $V$ and $a' + b' + c'$ by $V'$.\hfill\null

Then\hfill $V - V' = a$.\hfill (2)

We have $\text{Vol.\ } P < V$, and $\text{Vol.\ } P' > V$.\hfill §~\Sref{244}

\null\hfill $\therefore\quad \text{Vol.\ } P - \text{Vol. \ } P' < V - V'$.\hfill Why?

Hence from (2)\qquad $\text{Vol.\ } P - \text{Vol.\ } P' < a$.

Now take the divisions on $PM$ small enough to make $a < K$. \hfill §~\Sref{243}

Hence,\hfill $\text{Vol.\ } P - \text{Vol. \ } P' < K$.\hfill (3)

Thus (3) contradicts~(1), and hence the supposition that $P$
and $P'$ differ in volume is impossible.
\QED

\NewFolio %% -----File: 095.png---Folio 87-------

\Section{VOLUME OF A TRIANGULAR PYRAMID}

\begin{Theorem}{246.}{IV.}
The volume of a triangular
pyramid is one third of the product of its base and
altitude.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{095ab}{}
\end{center}

\Objective{Given the triangular pyramid $E-ABC$. Let $h$, $b$, and $V$ be the
numerical measures respectively of the altitude $EM$, the area of the
base $ABC$, and the volume of the pyramid.}
{To prove that\qquad $V = \tfrac{1}{3} bh$.}

\Proof: On the base $ABC$ construct a triangular prism with
altitude~$h$ and lateral edge~$EB$.

This prism may be cut into three pyramids by the plane
sections through $DEC$ and~$AEC$, as shown in the figure on
the right. % [**PP: Was ``...figure to the right'']

The pyramids $E-ABC$ and $C-DEF$ have equal volumes
since they have equal bases, $ABC$ and~$DEF$ (§~\Sref{175}), and the
same altitude,~$EM$.

Likewise volume $E-ACD = {}$ volume $E-CFD$.

But $C-DEF$ and $E-CFD$ are the same pyramid.

Hence, $\text{Vol.\ } E-ABC = \text{Vol.\ } C-DEF = \text{Vol.\ } E-ACD$.

That is, $\text{Vol.\ } E-ABC$ is one third of the volume of the
prism.
\begin{flalign*}
&\indent\text{But}    &     \text{volume of prism}   &= bh. \tag*{§~\Sref{198}} && \\
&\indent\text{Hence,} & V = \text{volume of pyramid} &= \tfrac{1}{3} bh. && \tag*{\QED}
\end{flalign*}

\NewFolio %% -----File: 096.png---Folio 88-------

\Section{VOLUME OF ANY PYRAMID}

\begin{Theorem}{247.}{V.}
The volume of any pyramid is
one third of the product of its base and altitude.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{096abc}{}
\end{center}

\Objective{Given the pyramid $P-ABCDE$. Let $V$, $b$, and~$h$ be the numerical
measures respectively of the volume, base, and altitude.}
{To prove that\qquad $V = \tfrac{1}{3} bh$.}

\Proof: By means of diagonal planes such as $PAC$ and $PAD$,
divide the given pyramid into triangular pyramids. Call the
bases of the triangular pyramids $b_1$, $b_2$, $b_3$, etc., and their common
altitude~$h$. Then by §~\Sref{246}, the volumes of the triangular
pyramids are $\tfrac{1}{3} b_1 h$, $\tfrac{1}{3} b_2 h$, $\tfrac{1}{3} b_3 h$.

Hence the total volume is $\tfrac{1}{3}(b_1 + b_2 + b_3)h = \tfrac{1}{3} bh$.
\QED

\begin{Corollary}{248.}{1.}
The volumes of any two pyramids having
equal altitudes are proportional to the areas of their bases.
\end{Corollary}

% [** Illustrations]
\begin{center}
\Input{096d_g}{\textwidth}
\end{center}

\begin{Corollary}{249.}{2.}
The volumes of any two pyramids having
bases of equal areas are proportional to their altitudes.
\end{Corollary}

\NewFolio %% -----File: 097.png---Folio 89-------

\Section{VOLUME OF A FRUSTUM OF A PYRAMID}

\begin{Theorem}{250.}{VI.}
The volume of a frustum of a
pyramid is equal to the combined volumes of three
pyramids whose common altitude is the same as that
of the frustum, and the areas of whose bases are those
of the upper and lower bases of the frustum and the
mean proportional between these areas.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{097ab}{}
\end{center}

\Objective{Given the frustum $AC'$ with area of lower base~$b$, area of upper
base~$b'$, and altitude~$h$. Then $\sqrt{bb'}$ is the mean proportional between
$b$ and $b'$.}
{To prove that the volume of $AC'$ is
\[
V = \tfrac{1}{3} hb + \tfrac{1}{3} hb' + \tfrac{1}{3} h \sqrt{bb'} = \tfrac{1}{3} h [b + b' + \sqrt{bb'}].
\]}

\Proof: Let $h'$ be the altitude $PK$ of the complete pyramid
$P-ABCDE$. Then the altitude of the pyramid $P-A'B'C'D'E'$
is $h' - h$.
\begin{flalign*}
&\indent\text{Hence,} & \dfrac{b}{b'} &= \dfrac{h'^2}{(h' - h)^2} && \tag*{§~\Sref{241}~(3)} \\
&\text{from which} & \dfrac{\sqrt{b}}{\sqrt{b'}} &= \dfrac{h'}{h' - h}. && \\
\intertext{\indent Clearing of fractions and solving for $h'$,}
&& h' &= \dfrac{h\sqrt{b}}{\sqrt{b} - \sqrt{b'}}. && \tag{1} % [** PP: Period added]
\end{flalign*}

\NewFolio %% -----File: 098.png---Folio 90-------

Now $V$ is the difference between the pyramids whose
altitudes are $h'$ and $h' - h$.
\begin{flalign*}
&\indent\text{Hence,}    & V &= \tfrac{1}{3} bh' - \tfrac{1}{3} b'(h' - h), && \tag*{§~\Sref{247}} \\
&\text{or, rearranging,} & V &= \tfrac{1}{3} b'h + \tfrac{1}{3} h'(b - b'). && \tag*{(2)}
\end{flalign*}

Substituting the value of $h'$ from~(1) in~(2) we have
\begin{align*}
V &= \dfrac{1}{3} b'h + \dfrac{1}{3} \dfrac{h\sqrt{b}}{\sqrt{b} - \sqrt{b'}} (b - b') \\
  &= \dfrac{1}{3} b'h + \dfrac{1}{3} \dfrac{h\sqrt{b}}{\sqrt{b} - \sqrt{b'}} (\sqrt{b} - \sqrt{b'}) (\sqrt{b} + \sqrt{b'}) \\
  &= \tfrac{1}{3} b'h + \tfrac{1}{3} hb + \tfrac{1}{3} h\sqrt{bb'} \\
  &= \tfrac{1}{3} h [b' + b + \sqrt{bb'}]. \tag*{\QED}
\end{align*}

\begin{Sightwork}

\item[1.] A flower bed is in the form of a regular right pyramid, with a
square base $5$~ft.\ on a side. The altitude is $3$~ft. Find the number of
cubic feet of soil used in its construction.

\item[2.] The altitude of a certain pyramid is $15$~in.\ and its volume is $380$~cu.~in.
Find the area of its base.

\item[3.] The area of the base of a pyramid is $48$~sq.~ft.\ and its volume
$160$~cu.~ft. Find its altitude.

\item[4.] Find the locus of the vertices of pyramids having the same base
and equal volumes.
\index{Loci@\textbf{Loci problems}}%

\item[5.] Two monuments having bases of equal areas are pyramidal in shape,
one being $15$~ft.\ high and the other $18$~ft. Find the ratio of their volumes.

\item[6.] Two pyramids with equal altitudes have bases whose areas are
$7$~sq.~ft.\ and $13$~sq.~ft. Find the ratio of the volumes of the pyramids.

\item[7.] Find the volume of a frustum of a pyramid the areas of whose bases
are $36$~sq.~in.\ and $144$~sq.~in., and whose altitude is $10$~in.

\item[8.] The volume of a frustum of a pyramid is $332$~cu.~in.\ and the areas
of its bases are $9$~sq.~in.\ and $36$~sq.~in. Find its altitude.
\end{Sightwork}

\NewFolio %% -----File: 099.png---Folio 91-------

\Section{EXERCISES INVOLVING NUMERICAL COMPUTATION}

\begin{Itemize}
\index{Problems@\textbf{Problems and Applications}|(}

\item[1.] The base of a regular pyramid is a square whose sides
are $16$~ft. Find the slant height of the pyramid if its altitude
is $6$~ft. Also find the lateral area.

\item[2.] The lateral area of a regular hexagonal pyramid is
$72$~sq.~ft.\ and the slant height is $12$~ft. Find the perimeter
of the base, the apothem of the base, and the altitude of the
pyramid.

\item[3.] Find the lateral area of the frustum of a pyramid if the
perimeters of its bases are $27$ and $54$~in.\ respectively and its
slant height is $12$~in. Does the result depend upon the number
of sides of the frustum?

\item[4.] The bases of a frustum of a pyramid are squares whose
sides are $8$~in.\ and $2$~in.\ respectively. Find the volume of the
frustum if its altitude is $6$~in.

\item[5.] Find the volume of a regular triangular pyramid the
sides of whose base are $5$~in.\ and whose altitude is $6$~in.

\begin{Suggestion}
The area of a regular triangle with side~$a$ is $\dfrac{a^2}{4}\sqrt{3}$.
\end{Suggestion}

\item[6.] Find the volume of a regular hexagonal pyramid the
sides of whose base are $8$~in.\ and whose altitude is $10$~in.

\begin{Suggestion}
The area of a regular hexagon with side~$a$ is $\dfrac{3 a^2}{2}\sqrt{3}$.
\end{Suggestion}

\item[7.] Two marble ornaments of equal altitudes are pyramidal
in form. One has a square base $2$~in.\ on a side and the other
a regular hexagonal base $1$~in.\ on a side. Find the ratio of
their volumes.

\item[8.] A pyramid has for its base a right triangle with hypotenuse
$10$ and shortest side~$6$. Find the volume of the pyramid
if its altitude is~$9$.

\item[9.] The slant height of a frustum of a regular pyramid is
$10$~in.\ and the apothems of its bases are $8$~in.\ and $6$~in.\ respectively.
Find its altitude.
\index{Problems@\textbf{Problems and Applications}|)}
\end{Itemize}

\NewFolio %% -----File: 100.png---Folio 92-------

\Section{EXERCISES INVOLVING ALGEBRAIC COMPUTATION}

\begin{Itemize}

\item[1.] A tent is to be made in the form of a right pyramid
with a regular hexagonal base. If the altitude is fixed at
$12$~ft., what must be the side of the base in order that the
tent may inclose $400$~cu.~ft.\ of space?

\item[2.] A pyramid with altitude $8$~in.\ and a base whose area is
$36$~sq.~in.\ is cut by a plane parallel to the base so that the
area of the section is $18$~sq.~in. Find the distance from the
base to the cutting plane.

\item[3.] A frustum of a pyramid is cut from a pyramid the
perimeter of whose base is $60$~in.\ and whose altitude is $15$~in.
What is the altitude of the frustum if the perimeter of its
upper base is $20$~in.? Does the result depend upon the number
of sides of the pyramid?

\item[4.] Solve the preceding problem if the perimeter of the
upper base of the frustum is one~$n$th that of the lower base.

\item[5.] The area of the base of a pyramid is $180$~sq.~in.\ and its
altitude is $20$~in. Cut from it a frustum the area of whose
upper base is $45$~sq.~in.; also one the area of whose upper base
is one~$n$th of $180$~sq.~in. Do these results depend upon the
number of sides of the pyramid?

\item[6.] If the altitude of a pyramid is $h$, how far from the base
must a plane parallel to it be drawn so that the area of its
cross-section shall be half that of the base of the pyramid?

\item[7.] In a regular right pyramid a plane parallel to the base
cuts it so as to make a section whose area is one half that of
the base. Find the ratio between the lateral area of the pyramid
and that of the small pyramid cut off by the plane.

\item[8.] Find the volume and the total surface of a regular tetrahedron
whose edges are $9$~in.

\item[9.] Find the total surface and the volume of a regular hexagonal
pyramid the sides of whose base are each~$a$ and whose
altitude is~$a$.
\end{Itemize}

\NewFolio %% -----File: 101.png---Folio 93-------

\Section{CONES}
\SecToc{Properties of Cones}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{251.}{Conical Surface.} Given a closed convex
\index{Cone@\textbf{Cone}!element of}%{\Pnum{251}}%
\index{Cone@\textbf{Cone}!vertex of}%{\Pnum{251}}%
\index{Conical@\textbf{Conical surface}!element of}%{\Pnum{251}}%
\index{Conical@\textbf{Conical surface}!generator of}%{\Pnum{251}}%
\index{Conical@\textbf{Conical surface}!vertex of}%{\Pnum{251}}%
\index{Element@\textbf{Element}!of a cone}%{\Pnum{251}}%
\index{Generator@\textbf{Generator}!of a conical surface}%{\Pnum{251}}%
\index{Surface@\textbf{Surface}!conical}%{\Pnum{251}}%
\index{Vertex@\textbf{Vertex}!of a cone}%{\Pnum{251}}%
plane curve and a fixed point $P$ not in its
plane. If a line through $P$ moves so as
always to touch the curve and is made to
traverse it completely, it is said to generate
a \emph{convex conical surface}.

\Indent The moving line is the \emph{generator} of the surface,
and in any of its positions it is an element
of the surface. The fixed point is the \emph{vertex}.

\Indent\Paragraph{252.}{Nappes.} A conical surface has two
\index{Conical@\textbf{Conical surface}!nappes of}%{\Pnum{252}}%
\index{Nappes@\textbf{Nappes}!of a conical surface}%{\Pnum{252}}%
parts, one on each side of the vertex, which are called \emph{nappes}.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{101a}{1.25in}}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{253.}{Cone.} The solid bounded by a conical surface and a
\index{Cone@\textbf{Cone}!altitude of}%{\Pnum{253}}%
\index{Section@\textbf{Section}!of a cone}%{\Pnum{253}}%
plane section cutting all its elements, and
not passing through the vertex, is called a
\emph{cone}.

\Indent\Paragraph{254.}{Base. Lateral Surface. Altitude.} The
plane part of the surface of a cone is its
\emph{base} and the curved part is the \emph{lateral surface}.
The \emph{altitude} of a cone
\index{Altitude@\textbf{Altitude}!of a cone}%{\Pnum{254}}%
\index{Base@\textbf{Base}!of a cone}%{\Pnum{254}}%
\index{Cone@\textbf{Cone}!altitude of}%{\Pnum{254}}%
\index{Cone@\textbf{Cone}!base of}%{\Pnum{254}}%
\index{Cone@\textbf{Cone}!lateral surface of}%{\Pnum{254}}%
is the perpendicular
distance from the vertex to the plane
of the base.
\end{minipage}%
% [** Illustration]
\Input{101b}{1.25in}

\Paragraph{255.}{Circular Cone.} A cone which has a circular cross-section
such that the perpendicular upon it from the vertex meets it
at the center is called a \emph{circular cone}. If the base is such a
circle, the cone is then a \emph{right circular cone}. Otherwise, it is
an \emph{oblique circular cone}.

\index{Axis@\textbf{Axis}!of a cone}%{\Pnum{255}}%
\index{Circular@\textbf{Circular}!cone}%{\Pnum{255}}%
\index{Cone@\textbf{Cone}!circular}%{\Pnum{255}}%
\index{Cone@\textbf{Cone}!oblique}%{\Pnum{255}}%
\index{Cone@\textbf{Cone}!right circular}%{\Pnum{255}}%
\index{Right@\textbf{Right}!cone}%{\Pnum{255}}%
The \emph{axis} of a circular cone is the line from the vertex
through the center of a circular section.

\begin{Sightwork}

\item[] If a plane cuts both nappes of a conical surface, show that it cannot
cut all the elements of the surface. See \textit{suggestion} on page~\Pgref{79}.
\end{Sightwork}

\NewFolio %% -----File: 102.png---Folio 94-------

\Paragraph{256.}{Generating a Right Circular Cone.} A right circular cone
\index{Cone@\textbf{Cone}!right circular}%{\Pnum{256}}%
may be \emph{generated} by rotating a right triangle $PMB$ about one
of its legs, $PM$, as an axis. The hypotenuse $PB$ generates the
lateral surface, and the other leg, $MB$,
generates the base.

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{257.}{Slant Height.} The generator
of the convex surface of a right circular
cone in any of its positions is
called the \emph{slant height} of the cone.
\index{Slant@\textbf{Slant height}!of a right cone}%{\Pnum{257}}%

\Indent\Paragraph{258.}{Frustum of a Cone.} The part of a cone included between
\index{Frustum@\textbf{Frustum}!of a cone}%{\Pnum{258}}%
the base and a plane
section parallel to the base is
called a \emph{frustum of a cone}.
\end{minipage}%
% [** Illustrations]
\hfill\Input{102ab}{2in}

The base of the cone and the
parallel section are the \emph{bases of
the frustum}.

\noindent\begin{minipage}[b]{\textwidth-2.25in}
\Indent\Paragraph{259.}{Slant Height of a Frustum.} A frustum cut from a right
\index{Slant@\textbf{Slant height}!of a frustum}%{\Pnum{259}}%
circular cone has two circular bases. The segments of all the
elements of the cone intercepted between these bases are equal
and their common length is the \emph{slant height} of the frustum.
\end{minipage}%
% [** Illustrations]
\hfill\Input{102cd}{2.25in}

\begin{Sightwork}

\item[1.] Compare the definitions of pyramidal and conical surfaces.

\item[2.] Compare the definitions of a pyramid and a cone, and of a frustum
of each.

\item[3.] What kind of cone corresponds to a regular pyramid?

\item[4.] Can the expression ``slant height'' be applied to any other cone
than a right circular cone? Discuss this question for the frustum of a
cone.

\item[5.] In a right circular cone, if any two of the three quantities, altitude,
slant height, radius of the base, are given, show that the third may be
found.
\end{Sightwork}

\NewFolio %% -----File: 103.png---Folio 95-------

\Section{SECTION OF A CONE THROUGH AN ELEMENT}

\begin{Theorem}{260.}{VII.}
If a plane contains an element
of a cone and meets its surface in any other point, then
it contains another element also, and the section is a
triangle.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{103ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Let a plane contain the element~$PB$ of the cone $P-ABC$, and also
one other point $K$ or~$K'$.}
{To prove that this plane contains another element~$PD$, and that
the section is a triangle~$PBD$.}

\textbf{Outline of Proof}: \Case{(1)}{When the point~$K$ is in the lateral surface.}

Draw the element $PD$ through $K$, and let the plane determined
by $PB$ and~$PD$ meet the base in~$BD$.

Then $\triangle PBD$ is the section made by the plane containing $PB$
and the point~$K$.

\Case{(2)}{When the point $K'$ is in the base.}

Draw $BD$ through $K'$ and also the element $PD$.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent Then the plane determined by $BD$
and $PD$ contains $PB$ and $K'$ and cuts the
cone in the $\triangle PBD$.

\Indent\Paragraph{261.}{Plane Tangent to a Cone.} If a
\index{Tangent@\textbf{Tangent}!to a cone}%{\Pnum{261}}%
plane contains an element of a cone and no
other point of the cone, the plane is \emph{tangent}
to the cone, and the element is called
the \emph{element of contact}.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{103c}{1.25in}}

\NewFolio %% -----File: 104.png---Folio 96-------

\Section{SECTION OF A CONE PARALLEL TO THE BASE}

\begin{Theorem}{262.}{VIII.}
If the base of a cone is circular,
every plane section parallel to the base is also circular.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{104ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given a cone with a circular base~$AD$.}
{To prove that the $\parallel$ section $EH$ is also circular.}

\Proof: Draw the straight line from $P$ to the center $M$ of the
base, and let it meet the section $EH$ in the point~$O$. Let $F$
and~$G$ be any two points on the perimeter of the section $EH$.

Pass planes containing $PM$ through the points $F$ and~$G$, and
let them cut the base in $MB$ and~$MC$ respectively.

Now in $\triangles PMB$ and $PMC$, $OF\parallel MB$ and $OG\parallel MC$.\hfill §~\Sref{92}
\[
\therefore\quad \dfrac{OF}{MB} = \dfrac{OP}{MP} = \dfrac{OG}{MC}. % [** PP: Added period]
\]

But\hfill $MB = MC$.\quad $\therefore\ OF = OG$.\hfill Why?

Hence, since $F$ and $G$, \emph{any} two points on the perimeter of
this section, are equally distant from~$O$, this shows that $EH$
is a circle whose center is~$O$.
\QED

\begin{Corollary}{263.}{}
If a cone has a circular base, the areas of
two cross-sections parallel to it are in the same ratio as the squares
of their perpendicular distances from the vertex and also as the
squares of the distances of their centers from the vertex.
\end{Corollary}

\NewFolio %% -----File: 105.png---Folio 97-------

\Section{INSCRIBED AND CIRCUMSCRIBED PYRAMIDS}

\Paragraph{264.}{} The lateral surface of a regular pyramid inscribed in a
\index{Circumscribed@\textbf{Circumscribed}!cone}%{\Pnum{264}}%
\index{Circumscribed@\textbf{Circumscribed}!pyramid}%{\Pnum{264}}%
\index{Inscribed@\textbf{Inscribed}!cone}%{\Pnum{264}}%
\index{Inscribed@\textbf{Inscribed}!pyramid}%{\Pnum{264}}%
right circular cone may be computed and is equal to one half
the product of the slant height and the perimeter
of the base (§~\Sref{237}).
\Page{96}%

\noindent\begin{minipage}[b]{\textwidth-1in}
\Indent If the number of faces of the inscribed pyramid
is doubled, the lateral surface of the resulting
pyramid may again be computed in terms of its
slant height and the perimeter of its base.
\end{minipage}%
% [** Illustration]
\hfill\raisebox{-0.125in}{\smash{\Input{105a}{1in}}}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent In a similar manner, the lateral surface of a
regular pyramid circumscribed about a right
circular cone may be computed in terms of the
slant height of the cone and the perimeter of
the polygon circumscribed about the base.
The number of faces may be doubled and the
lateral surface again computed, and so on.
\end{minipage}%
% [** Illustration]
\hfill\Input{105b}{1.25in}

Evidently either of these processes may be repeated indefinitely
and the surfaces of the inscribed or circumscribed
pyramids may be made to lie as close to the surface of the cone
as we please.

\begin{Remark}
The circumscribed pyramids all have the same slant height as that of
the cone, and in case of the inscribed pyramids, the slant height may be
made to differ by as little as we please from that of the cone by making
the number of faces great enough.
\end{Remark}

\Paragraph{265.}{Fundamental Assumption on the Lateral Area of a Right
Circular Cone.} We now assume that
\index{Area@\textbf{Area}!of a cone}%{\Pnum{265}}%
\index{Axioms@\textbf{Axioms}}%{\Pnum{265}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{265}}%

\textit{A right circular cone has a definite lateral area which can be
approximated as nearly as we please by taking the lateral area of
the successive inscribed or circumscribed pyramids}.

\begin{Remark}
Since, by the theorem of §~\Sref{237}, the lateral area of any regular pyramid
is half the product of the perimeter of its base and its slant height, it
follows that this theorem holds for all pyramids used in approximating
the lateral area of a right circular cone and that all practical measurements
of such lateral areas are based on this theorem.
\end{Remark}

\NewFolio %% -----File: 106.png---Folio 98-------

\Section{LATERAL AREA OF A RIGHT CIRCULAR CONE}

The foregoing considerations constitute an informal proof
of the following theorem:

\begin{Theorem}{266.}{IX.}
\index{Cone@\textbf{Cone}!lateral surface of}%{\Pnum{266}}%
The area of the lateral surface
of a right circular cone is equal to one half the product
of its slant height and the circumference of its base.
\end{Theorem}

The argument used on page~\Pgref{96} shows that a theorem of this
kind holds for every inscribed or circumscribed pyramid used in
the approximation process, and hence this theorem for the cone
is established for all purposes of practical measurement.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\begin{Corollary}{267.}{}
If $l$ is the slant height
of a right circular cone and $r$ is the radius of
its base, the area of the lateral surface is
\[
S = \tfrac{1}{2}\cdot 2 \pi r\cdot l = \pi r l.
\]
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.25in}{\smash{\Input{106a}{2in}}}

\textit{Note}. In the case of a cone which is not a right circular cone the
slant height varies from point to point and the process of computation of
§~\Sref{266} fails. Finding the lateral surface of such a cone depends on
methods first introduced in the calculus and is a much more difficult
problem than those solved in elementary plane and solid geometry.

\begin{Sightwork}

\item[1.] If a pyramid is inscribed in a cone, is its lateral area greater or less
than that of the cone?

\item[2.] Does the theorem of §~\Sref{237} apply to an irregular pyramid or to an
oblique pyramid? Discuss fully.

\item[3.] If a regular pyramid of $1000$~faces is inscribed in a right circular
cone, would the theorem of §~\Sref{237} apply to this pyramid? Would it be
easy to detect experimentally the difference between this pyramid and the
cone? Would the size of the cone affect the answer to this question?
Discuss similar questions about a circumscribed pyramid of $1000$~faces.

\item[4.] Find the lateral surface of a right circular cone whose altitude is
$7$~in.\ and the radius of whose base is $2$~in.
\end{Sightwork}

\NewFolio %% -----File: 107.png---Folio 99-------

\Section{LATERAL AREA OF A FRUSTUM}

\begin{Theorem}{268.}{X.}
The lateral area of a frustum
of a right circular cone is equal to one half the sum
of the circumferences of the bases multiplied by the
slant height.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{107ab}{}
\end{center}

\Objective{Given the frustum $ABCD$, with slant height $l$ and radii of bases
$r$ and $r'$. \\
\indent Let $S$ represent its lateral area.}
{To prove that $S = \tfrac{1}{2} (2\pi r + 2\pi r')l = \pi l(r + r')$.}

\Proof: Complete the cone, and let $PC = l'$.
\begin{flalign*}
&\text{Then} &   S &= \tfrac{1}{2} \bigl[2\pi r(l + l') - 2\pi r'l'\bigr] && \tag*{§~\Sref{267}} \\
&&                 &= \pi rl + \pi l'(r - r'). && \tag*{(1)} \\
&\text{But} & \dfrac{r}{r'} &= \dfrac{l + l'}{l'}, \text{ from which } l' = \dfrac{r'l}{r - r'}. && \tag*{(2)} \\
\intertext{Substituting $l'$ from (2) in (1),}
&&               S &= \pi rl + \pi r'l = \pi l(r + r'). && \tag*{\QED}
\end{flalign*}

\begin{Corollary}{269.}{}
The lateral area of a frustum of a right
circular cone is equal to the circumference of a section midway
between the bases multiplied by the slant height.
\end{Corollary}

\begin{Suggestion}
In the formula of the theorem we have
\[
S = \pi l(r + r') = 2\pi \dfrac{(r + r')}{2} l.
\]

Now show that $\dfrac{r + r'}{2}$ is the radius of the section midway between the
two bases.\hfill §~\Sref{46}
\end{Suggestion}

\NewFolio %% -----File: 108.png---Folio 100-------

\begin{Sightwork}

\item[1.] If a triangle which is not a right triangle revolves about one of its
sides, does it generate a cone?

Show that the figure generated by revolving any triangle about its
longest side may be divided into two cones.

\item[2.] Find the lateral area of a right circular cone with radius of base $6$~in.\
and slant height $10$~in. In this and the succeeding exercises, express the
results in terms of~$\pi$, e.g., $60 \pi$~sq.~in.

\item[3.] Find the lateral area of a right circular cone with radius of base $8$~in.\
and altitude $6$~in.

\item[4.] Find the lateral area of a frustum of a right circular cone, the
radii of whose bases are $8$~in.\ and $4$~in.\ and whose slant height is $6$~in.

\item[5.] A right circular cone having a base with radius $6$~ft.\ and altitude
$8$~ft.\ is cut by a plane parallel to its base and at a distance of $4$~ft.\ from
the vertex. Find the radius of this section. Also find its area.

\end{Sightwork}

\begin{Exercises}

\item[1.] The lateral area of the surface of a right circular cone is
$120\pi$~sq.~in., and its radius is $4$~in. Find its slant height.

\item[2.] A circular chimney $100$~ft.\ high is in the form of a
frustum of a right circular cone whose lower base is $10$~ft.\ in
diameter and upper base $8$~ft. Find the lateral surface.

\item[3.] The lateral area of a frustum of a right circular cone is
$60\pi$~sq.~in.; the radii of the two bases are $6$~in.\ and $4$~in. Find
the slant height of the frustum.

\item[4.] The lateral area of a right circular cone is $S$, and the slant
height is~$l$. Express the radius of the base and also the altitude
in terms of $S$ and~$l$.

\item[5.] If the radius of the base of a right circular cone is $r$, and
the lateral area is $S$, express the slant height in terms of $r$
and~$S$.

\item[6.] If the slant height of a right circular cone is~$l$, and the
lateral area is~$S$, express the circumference of the base in
terms of $l$ and~$S$.
\end{Exercises}

\NewFolio %% -----File: 109.png---Folio 101-------

\Section{VOLUME OF A CONE}

\Paragraph{270.}{} Consider inscribed and circumscribed pyramids similar
\index{Cone@\textbf{Cone}!volume of}%{\Pnum{270}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{270}}%
to those used in §~\Sref{264}, except that they need not be regular
since the cone is not now required to be a right circular cone.

By repeatedly increasing the number of faces of either the
inscribed or circumscribed pyramids, they may be made to
approach coincidence with the cone as nearly as we please.

\Paragraph{271.}{Fundamental Assumption on the Volume of a Cone.} We
\index{Axioms@\textbf{Axioms}}%{\Pnum{271}}%
\index{Cone@\textbf{Cone}!volume of}%{\Pnum{271}}%
now assume that \textit{a cone has a definite volume which can be approximated
as nearly as we please by taking the volume of successive
inscribed or circumscribed prisms}.

Recalling that the volume of a pyramid is equal to one
third of the product of its base and altitude, we see that the
above considerations constitute an informal proof of the
following theorem:

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Theorem}{272.}{XI.}
\index{Volume@\textbf{Volume}!of a cone}%{\Pnum{272}}%
The volume of a
cone is equal to one third of the product
of its base and altitude.
\end{Theorem}

\Indent\begin{Corollary}{273.}{}
If the base of a cone is a
circle with radius~$r$ and if the altitude is~$h$, then
the volume of the cone is
\[
V = \tfrac{1}{3}\cdot \pi r^2\cdot h = \tfrac{1}{3} \pi r^2 h.
\]
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.375in}{\Input{109a}{1.25in}}

\begin{Sightwork}

\item[1.] Find the volume of an oblique cone with altitude $8$~in.\ and a circular
base whose radius is $6$~in.

\item[2.] Find the volume of a right circular cone with slant height $10$~in.\
and radius of base $6$~in.

\item[3.] The area of the base of a cone is $50$~sq.~in.\ and its volume is $600$~cu.~in.
Is the altitude the same whether the cone is right or oblique?

\item[4.] Show that if two cones have bases of equal areas their volumes are
proportional to their altitudes.
\end{Sightwork}

\NewFolio %% -----File: 110.png---Folio 102-------

\Section{VOLUME OF A FRUSTUM OF A CONE}

\begin{Theorem}{274.}{XII.}
The volume of the frustum of
a cone is equal to the combined volumes of three cones
whose common altitude is the altitude of the frustum
and whose bases are the upper and lower bases of the
frustum and a mean proportional between these bases.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{110ab}{}
\end{center}
\vspace*{-0.125in}

\begin{Suggestion}
The proof is exactly like that of §~\Sref{250}, making use of
§~\Sref{272}, instead of §~\Sref{247}. The result in symbols is $V = \tfrac{1}{3} h(b + b' + \sqrt{bb'})$.
\end{Suggestion}

\begin{Corollary}{275.}{}
The volume of a frustum of a right circular
cone is $V = \tfrac{1}{3} \pi h(r^2 + {r'}^2 + rr')$, where $h$ is the altitude and
$r$ and~$r'$ are the radii of the bases.
\end{Corollary}
\begin{flalign*}
& \text{For } & b = \pi r^2,\ b' = \pi {r'}^2.\quad \therefore\
V &= \tfrac{1}{3} h(\pi r^2 + \pi {r'}^2 + \sqrt{\pi r^2\cdot \pi {r'}^2}) && && \\
&&  &= \tfrac{1}{3} \pi h(r^2 + {r'}^2 + rr'). && &&
\end{flalign*}

\textit{Note}. The theorem of §~\Sref{272} holds for any cone whatever, whether
right or oblique. However, when the base is not a circle we have no
means in elementary mathematics for \emph{computing} its area. Hence the
volume of such a cone cannot be found at this stage even though §~\Sref{272}
does apply.

Similarly §~\Sref{274} applies to a frustum of any cone whatever, but we are
able to \emph{compute} its volume by elementary methods only in case the bases
are circles.

\begin{Sightwork}

\item[] The radius of the base of a cone is $5$~in.\ and its altitude is $10$~in. Find
the volume of a frustum formed by a plane parallel to the base and $6$~in.\
from it. Find the total surface of this frustum in case the cone is a right
circular cone.
\end{Sightwork}

\NewFolio %% -----File: 111.png---Folio 103-------

\Section{SUMMARY OF BOOK IV}
\index{Summaries@\textbf{Summaries}}

\begin{Itemize}

\item[1.] Define pyramidal surface and conical surface. In what
respects do they differ?

\item[2.] Define pyramid, cone, regular pyramid, circular cone,
and right circular cone.

\item[3.] For what kind of cones may the lateral area be found
by means given in this Book? What is the corresponding
kind of pyramid?

\item[4.] For what kind of frustum of a cone may the lateral
area be found by means given in this Book? What is the
corresponding kind of frustum of a pyramid?

\item[5.] What assumption about the volume of a pyramid is
made in Book~IV? In the proof of what theorems is this
assumption used?

\item[6.] Beginning with the theorem of §~\Sref{245} state in order the
theorems which lead to the rule for finding the volume of any
pyramid.

\item[7.] What assumption about the lateral area of a cone is
made in this Book? Compare this assumption with the one
in §~\Sref{219}.

\item[8.] What assumption is made about the volume of a cone?
Compare this with the assumption in §~\Sref{219}.

\item[9.] What theorems on cylinders have no corresponding
theorems for cones?

\item[10.] Show that a frustum of a cone becomes more and more
nearly identical with a cylinder if the vertex of the cone is
removed farther and farther from the base.

\item[11.] If a cone were regarded as a pyramid with a very large
number of very narrow lateral faces, what pages in this Book
would be superfluous?

\item[12.] Describe the difficulty in finding the lateral area of an
oblique cone or of a cone whose base is not circular. Does the
same difficulty exist in finding the volume of such a cone?
\end{Itemize}

\NewFolio %% -----File: 112.png---Folio 104-------

\Section{EXERCISES ON BOOK IV}

\begin{Itemize}

\item[1.] If several planes are tangent to the same cone, find one
point common to them all.

\item[2.] Find the locus of all lines which make a given angle with
a given line at a given point in it.
\index{Loci@\textbf{Loci problems}}%

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[3.] Find the locus of all lines which make
a given angle with a given plane at a given
point.

% [** Illustration]

\Indent\item[4.] If the middle points of four edges of
a tetrahedron, no three of which meet at the
same vertex, are joined, prove that a parallelogram
is formed.

\Indent\item[5.] Show how to pass a plane through a
tetrahedron so that the section shall be a
parallelogram.

\Indent\begin{Suggestion}
Pass a plane parallel to each of two
opposite edges. See §~\Sref{133}.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{112ab}{1.25in}}


\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\item[6.] A mound of earth of the shape shown in the figure has
a rectangular base $16$~yards long and $8$~yards
wide. Its perpendicular height
is $5$~yards, and the length on top is
$8$~yards. Find the number of cubic
yards of earth in the mound.
\end{minipage}%
% [** Illustration]
\hfill\Input{112c}{1.75in}

\begin{Suggestion}
If from each end a pyramid with a base $8$~yd.\ by $4$~yd.\
is removed, the remaining part is a triangular prism.
\end{Suggestion}

\item[7.] Given a figure in general shape the same as the preceding,
with a rectangular base of length $24$~ft.\ and width $6$~ft.
Find its volume and lateral area if the dihedral angles
around the base are each~$45$°.

\item[8.] Find the area and volume of the figure developed by an
equilateral triangle with sides a if it is revolved about one of
its sides.

\begin{Suggestion}
The figure may be divided into two cones.
\end{Suggestion}
\end{Itemize}

\NewFolio %% -----File: 113.png---Folio 105-------

\begin{Itemize}

\item[9.] Find the volume and area of the figure formed by revolving
an equilateral triangle with sides s about an altitude.

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\item[10.] Find the area and volume of the figure developed
by revolving a square whose side is a
about one of its diagonals.

\Indent\item[11.] Through one vertex of an equilateral triangle
with sides a draw a line~$l$ perpendicular to
the altitude upon the opposite side. Find the
volume and area of the figure developed by revolving
the triangle about the line~$l$.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{113a}{1in}}

\begin{Suggestion}
The volume may be obtained by subtracting the volumes
of two cones from the volume of a cylinder.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent\item[12.] Through a vertex of a square with sides~$a$
draw a line~$l$ perpendicular to the diagonal at that
point. Find the area and volume of the figure
developed by turning the square around~$l$.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{113b}{1in}}

\begin{Suggestion}
Notice that two frustums and two cones are developed.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1.125in}
\Indent\item[13.] In a regular hexagon with sides~$a$ draw a line~$l$
bisecting two opposite sides. Find the area and
volume of the figure developed by turning the hexagon
about $l$ as an axis.

\Indent\item[14.] One angle of a right triangle is~$30$°. Find
the ratios of the surfaces and also of the volumes
of the solids developed by revolving this triangle around each
of its three sides in succession.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{113c}{1in}}

\begin{Suggestion}
The sides of the $\triangle$ are $a$, $2a$,
$a\sqrt{3}$.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-2.25in}
\Indent\item[15.] If through any point~$P$ in a diagonal
of a parallelopiped planes $KN$ and
$RM$ are drawn parallel to two faces,
show that the parallelopipeds $DQ$ and $LN$ thus formed have
equal volumes.
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.375in}{\Input{113d}{2in}}}

\end{Itemize}

\NewFolio %% -----File: 114.png---Folio 106-------

\begin{BioSketch}{Archimedes}{Archimedes}
(287--212 B.C.) was without doubt the greatest mathematician of his
time. He is known not through any extended treatise, like that of
Euclid, but through a series of monographs on practically every
mathematical subject then known, including physics, mechanics,
astronomy, and many phases of geometry. His work on the circle, cone,
cylinder, and sphere is reflected in our treatment of surfaces and
volumes at the present day.
\end{BioSketch}

\NewFolio %% -----File: 115.png---Folio 107-------

\Chapter{BOOK V}{The Sphere}

\Paragraph{276.}{Sphere. Center.} A \emph{sphere} is a solid bounded by a surface
\index{Center@\textbf{Center}!of a sphere}%{\Pnum{276}}%
\index{Sphere@\textbf{Sphere}}%{\Pnum{276}}%
\index{Sphere@\textbf{Sphere}!center of}%{\Pnum{276}}%
all points of which are equally distant from a point
within called the \emph{center}.

\Paragraph{277.}{Diameter. Radius.} A line-segment connecting two
points on the surface of a sphere and passing through its
center is a \emph{diameter}. A segment joining the center to a point
on the surface is a \emph{radius}.
\index{Radius@\textbf{Radius}!of a sphere}%{\Pnum{277}}%
\index{Sphere@\textbf{Sphere}!radius of}%{\Pnum{277}}%

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{278.}{Notation for a Sphere.} A sphere
may be designated by a single letter at
the center or by two letters giving a
radius.

\Indent\begin{Remark}
Thus the sphere $C$ means one whose center
is $C$, and the sphere $CA$ is one whose center
is $C$ and whose radius is $CA$.
\end{Remark}

\Indent\Paragraph{279.}{Generating a Sphere.} The surface
of a sphere may be \emph{generated} by revolving
a semicircle about its diameter as an
axis.
\index{Spherical@\textbf{Spherical}!surface}%{\Pnum{279}}%
\index{Surface@\textbf{Surface}!spherical}%{\Pnum{279}}%

% [** Illustration]

\Indent\begin{Remark}
Thus the surface of the sphere $CA$ may be
generated by revolving the semicircle $MNA$
about the diameter $MN$.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfil\Input{115ab}{1.5in}\hfil

\begin{Corollary}{280.}{1.}
If two spheres have equal radii they may
be made to coincide and hence are equal.
\end{Corollary}

\begin{Corollary}{281.}{2.}
All radii of the same sphere or of equal
spheres are equal.
\end{Corollary}

\begin{Corollary}{282.}{3.}
All diameters of the same sphere or of
equal spheres are equal.
\end{Corollary}

\NewFolio %% -----File: 116.png---Folio 108-------

\Section{PLANE SECTION OF A SPHERE}

\begin{Theorem}{283.}{I.}
\index{Section@\textbf{Section}!of a sphere}%{\Pnum{283}}%
A section of a spherical surface
made by a plane is a circle.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{116ab}{}
\end{center}

\Objective{Given a sphere with center $C$ cut by a plane~$M$.}
{To prove that the points common to the surface of the sphere
and the plane form a circle.}

\Proof: From the center $C$ draw $CA$ perpendicular to the
plane~$M$.

Let $B$ and $D$ be any two points common to the plane and
the surface of the sphere. Complete the figure and prove
$AB = AD$.

Hence any two points common to the surface of the sphere
and the plane~$M$ are equidistant from the point~$A$. That is,
these points lie on a circle.

How must this proof be modified in case the plane~$M$
passes through the center of the sphere?

\begin{Corollary}{284.}{1.}
Through three points on a spherical surface
there is one and only one circle.
\end{Corollary}

\begin{Suggestion}
How many planes pass through these points?
\end{Suggestion}

\begin{Corollary}{285.}{2.}
A radius of a sphere through the center
of a circle on its surface is perpendicular to the plane of the
circle; and conversely, a radius of a sphere perpendicular to the
plane of a circle on its surface passes through the center of
the circle.
\end{Corollary}

\NewFolio %% -----File: 117.png---Folio 109-------

\Section{PROPERTIES OF CIRCLES ON A SPHERE}

\Paragraph{286.}{Figures on a Sphere.} Any figure drawn on the surface
\index{Figures@\textbf{Figures}!on a sphere}%{\Pnum{286}}%
of a sphere is said to \emph{lie on the sphere}.

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\Paragraph{287.}{Axis and Poles.} The line perpendicular
\index{Axis@\textbf{Axis}!of a circle on a sphere}%{\Pnum{287}}%
\index{Circle@\textbf{Circle}!axis of}%{\Pnum{287}}%
\index{Circle@\textbf{Circle}!poles of}%{\Pnum{287}}%
\index{Great@\textbf{Great circle}!axis of}%{\Pnum{287}}%
\index{Great@\textbf{Great circle}!pole of}%{\Pnum{287}}%
\index{Poles@\textbf{Poles}!of a circle}%{\Pnum{287}}%
to the plane of a circle at its
center is called the \emph{axis} of the circle.

\Indent The points in which the axis of a
circle on a sphere meets the surface of
the sphere are called the \emph{poles of the
circle}.
\end{minipage}%
% [** Illustration]
\hfil\Input{117a}{1.5in}\hfil

\Paragraph{288.}{Great and Small Circles on a Sphere.} If the plane of a
\index{Circle@\textbf{Circle}!great}%{\Pnum{288}}%
\index{Circle@\textbf{Circle}!small}%{\Pnum{288}}%
\index{Great@\textbf{Great circle}!on a sphere}%{\Pnum{288}}%
\index{Small@\textbf{Small circle} on a sphere}%{\Pnum{288}}%
\index{Sphere@\textbf{Sphere}!great circle of}%{\Pnum{288}}%
\index{Sphere@\textbf{Sphere}!small circle of}%{\Pnum{288}}%
circle on a sphere passes through the center of the sphere, it is
called a \emph{great circle} of the sphere, and if not, it is called a \emph{small
circle}.

\begin{Remark}
Thus, in the figure, $AB$ is a small circle, $PP'$ is its axis, and $P$ and~$P'$
are its poles.

The circle passing through $P$ and~$P'$ and containing the center~$C$ is a
great circle.
\end{Remark}

\Paragraph{289.}{Inside and Outside of a Sphere.} A point is inside, outside,
or on a sphere according as its distance from the center is less
than, greater than, or equal to the radius of the sphere.
\index{Sphere@\textbf{Sphere}!points within, without}%{\Pnum{289}}%

\begin{Corollary}{290.}{1.}
The center of a great circle on a sphere is
the center of the sphere.
\end{Corollary}

\begin{Corollary}{291.}{2.}
All great circles on a sphere are equal
and any two such circles bisect each other.
\end{Corollary}

\begin{Corollary}{292.}{3.}
The axis of any circle on a sphere passes
through the center of the sphere.
\end{Corollary}

\begin{Corollary}{293.}{4.}
Through two given points on a sphere
there is one and only one great circle unless these points are at
opposite ends of a diameter.
\end{Corollary}

\begin{Corollary}{294.}{5.}
Circles on a sphere formed by parallel
planes have the same axis and the same poles.
\end{Corollary}

\NewFolio %% -----File: 118.png---Folio 110-------

\Section{DISTANCES FROM POINTS ON A CIRCLE TO ITS POLE}

\Paragraph{295.}{Distance on a Sphere.} The \emph{spherical distance}, or simply
\index{Arc@\textbf{Arc} of a great circle}%{\Pnum{295}}%
\index{Distance@\textbf{Distance}!between two points on a sphere}%{\Pnum{295}}%
the \emph{distance} between two points on a sphere is the distance
measured between these points along the minor arc of the great
circle through them.

\begin{Theorem}{296.}{II.}
All points of a circle on a sphere
are equidistant from either pole of the circle.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{118ab}{}
\end{center}

\Objective{Given $P$ a pole of the circle whose center is $A$, with $B$, $C$, and $D$
any points on this circle.}
{To prove that the great circle arcs $PB$, $PC$, and $PD$ are equal.}

\begin{Suggestion}
Prove chord $PB = {}$chord $PC = {}$chord $PD$, and hence
arc $PB = {}$arc $PC = {}$arc $PD$.\hfill §~\Sref{48}
\end{Suggestion}

\Paragraph{297.}{Polar Distance of a Circle.} The spherical distance from
\index{Distance@\textbf{Distance}!polar}%{\Pnum{297}}%
\index{Polar@\textbf{Polar}!distance}%{\Pnum{297}}%
the points of a circle on a sphere to its nearest pole is called
the \emph{polar distance} of the circle.

\Paragraph{298.}{A Quadrant.} One fourth of a great circle is a \emph{quadrant}.
\index{Quadrant@\textbf{Quadrant}}%{\Pnum{298}}%

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\begin{Corollary}{299.}{1.}
The polar distance
of a great circle is a quadrant.
\end{Corollary}

\Indent\begin{Corollary}{300.}{2.}
If a point on a
sphere is at a quadrant's distance from
each of two points not at the extremities of
the same diameter, it is the pole of the
great circle through these points.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfill\Input{118c}{1.5in}

\NewFolio %% -----File: 119.png---Folio 111-------

\Section{USE OF A SPHERICAL BLACKBOARD}

\begin{Corollary}{301.}{3.}
Polar distances of equal circles on a
sphere are equal.
\end{Corollary}

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\begin{Corollary}{302.}{4.}
\index{Spherical@\textbf{Spherical}!blackboard}%{\Pnum{302}}%
Straight line-segments joining points of a
circle on a sphere, to one of its poles are
equal.
\end{Corollary}

\Indent\begin{Remark}
It follows from the preceding theorem and
corollaries that, if a \emph{spherical} blackboard is
at hand, circles may be constructed on it by
means of crayon and string the same as on
a \emph{plane} blackboard. Likewise, curve-legged
compasses may be used.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfil\Input{119a}{1.75in}\hfil

\begin{Sightwork}

\item[1.] If two points are at the extremities of the same diameter of a
sphere, how many great circles can be passed through these points?

\item[2.] What great circles on the earth's surface pass through both poles?
Where are the poles of these circles located?

\item[3.] If $A$ and $B$ are at opposite ends of a diameter, can a small circle be
passed through them?

\item[4.] If two circles on a sphere have the same poles, prove that their
planes are parallel. See §~\Sref{294}.

\item[5.] What is the locus of all points on a sphere at a quadrant's distance
from a given point?
\index{Loci@\textbf{Loci problems}|(}%

\item[6.] What is the locus of all points on a sphere at any fixed distance
from a given point on the sphere? What is the greatest such distance
possible? Discuss fully.

\item[7.] If two planes cutting a sphere are parallel, what can be said of the
centers of the circles thus formed? What can be said of the poles of these
circles? Are these statements true of three or more such circles?

\item[8.] Find the locus of the centers of a set of circles on a sphere formed
by a set of parallel planes cutting it.

\item[9.] $AB$ is a fixed diameter of a sphere. A plane containing $AB$ is
made to revolve about it as an axis. Find the locus of the poles of the
great circles on the sphere made by this revolving plane. How are the
points $A$ and $B$ related to this locus?
\index{Loci@\textbf{Loci problems}|)}%
\end{Sightwork}

\NewFolio %% -----File: 120.png---Folio 112-------

\Section{CIRCLES EQUIDISTANT FROM THE CENTER}

\begin{Theorem}{303.}{III.}
If the planes of two circles on
a sphere are equidistant from the center, the circles are
equal; and conversely, if two circles on a sphere are
equal, their planes are equidistant from the center.
\end{Theorem}

% [** Illustrations] FIG. 1   FIG. 2
\begin{center}
\Input{120ab}{}
\end{center}

\begin{Suggestion}
In the first figure show that (1)~if $CA = CA'$, then
$AB = A'B'$, and (2)~if $AB = A'B'$, then $CA = CA'$.
\end{Suggestion}

\Section{CIRCLES UNEQUALLY DISTANT FROM THE CENTER}

\begin{Theorem}{304.}{IV.}
If the planes of two circles on a
sphere are unequally distant from the center, the circles
are unequal, the one nearer the center being the greater;
and conversely, if two circles on a sphere are unequal,
the plane of the greater circle is nearer the center.
\end{Theorem}

\begin{Suggestion}
In \textsc{Fig}.~2 above, show (1)~that if $CA < CA'$ then
$AB > A'B'$; and (2)~if $AB > A'B'$ then $CA < CA'$.
\end{Suggestion}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{305.}{Plane Tangent to a Sphere.} A
plane which meets a sphere in only
one point is \emph{tangent to the sphere}.
\index{Sphere@\textbf{Sphere}!tangent to}%{\Pnum{305}}%
\index{Tangent@\textbf{Tangent}!to a sphere}%{\Pnum{305}}%

\Indent Two spheres are \emph{tangent to each other}
if they are both tangent to the same
plane at the same point.
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.25in}{\smash{\Input{120c}{1.5in}}}

A \emph{line is tangent to a sphere} if it contains one and only one
point of the sphere.

\NewFolio %% -----File: 121.png---Folio 113-------

\Section{PLANE TANGENT TO A SPHERE}

\begin{Theorem}{306.}{V.}
A plane tangent to a sphere is
perpendicular to the radius from the point of tangency;
and conversely, a plane perpendicular to a radius at its
extremity is tangent to the sphere.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{121ab}{}
\end{center}

\Objective{Given a sphere $C$ with plane $M$ tangent to it at $A$.}
{To prove that the radius $CA$ is perpendicular to the plane~$M$.}

\Proof: (1)~It is necessary to prove that $CA$ is perpendicular
to every line in $M$ through~$A$. (Why?)

Draw any such line as~$AB$.

The plane $BAC$ cuts the sphere in a circle. Then $AB$ is
tangent to this circle, and hence perpendicular to $AC$.

Since $CA$ is $\perp$ to every line in~$M$ through~$A$, it is $\perp$ to~$M$.

(2)~To prove the converse, note that $CA$ is the shortest
distance from $C$ to the plane~$M$.\hfill §~\Sref{82}

Hence, every point of $M$ except $A$ is exterior to the sphere.

That is, $M$ is a tangent plane. (§~\Sref{305}.)
\QED

\Paragraph{307.}{Sphere Inscribed in a Polyhedron.} A sphere is said to
\index{Circumscribed@\textbf{Circumscribed}!polyhedron}%{\Pnum{307}}%
\index{Inscribed@\textbf{Inscribed}!sphere}%{\Pnum{307}}%
\index{Polyhedrons@\textbf{Polyhedrons}!circumscribed}%{\Pnum{307}}%
\index{Sphere@\textbf{Sphere}!inscribed}%{\Pnum{307}}%
be \emph{inscribed in a polyhedron} if every face of the polyhedron is
tangent to the sphere. The polyhedron is then said to be \emph{circumscribed
about the sphere}.

\Paragraph{308.}{Polyhedron Inscribed in a Sphere.} A polyhedron is \emph{inscribed
in a sphere} if all its vertices lie in the surface of the
sphere. The sphere is then \emph{circumscribed about the polyhedron}.
\index{Inscribed@\textbf{Inscribed}!polyhedron}%{\Pnum{217}}%
\index{Polyhedrons@\textbf{Polyhedrons}!inscribed}%{\Pnum{308}}%

\NewFolio %% -----File: 122.png---Folio 114-------

\Section{SPHERE INSCRIBED IN A TETRAHEDRON}

\begin{Problem}{309.}
\index{Tetrahedron@\textbf{Tetrahedron}!circumscribed}%{\Pnum{309}}%
To inscribe a sphere in a given tetrahedron.
\end{Problem}

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{122ab}{\textwidth}
\end{center}
\vspace*{-0.25in}

\Objective{Given a tetrahedron $D-ABC$.}
{To construct a sphere tangent to each of its four faces.}

\begin{Construction}
Construct planes bisecting the dihedral angles
whose edges are $AB$, $BC$, and $CA$. These three planes meet in
a point~$P$, which is equally distant from the four faces of the
tetrahedron. With the point~$P$ as a center and a radius~$PE$
equal to the distance from $P$ to one of the faces construct a
sphere. This sphere is inscribed in the tetrahedron.
\end{Construction}

\Proof: Every point in the plane $PAB$ is equidistant from the
faces $ABD$ and $ABC$.\hfill §~\Sref{123}

Similarly every point in the plane $PBC$ is equidistant from
the planes $ABC$ and $DBC$, and every point in the plane $PAC$
is equidistant from the planes $ABC$ and $DAC$.

Hence, the point $P$, common to all three planes, is equidistant
from each of the four faces of the tetrahedron.

$\therefore$ each plane is tangent to the sphere, and the sphere is
inscribed in the tetrahedron. (§~\Sref{307}.)
\QEF

\begin{Sightwork}

\item[] Discuss the problem of finding a point equally distant from three planes
two of which are parallel. How many such points are there?
\end{Sightwork}

\NewFolio %% -----File: 123.png---Folio 115-------

\Section{SPHERE CIRCUMSCRIBED ABOUT A TETRAHEDRON}

\begin{Problem}{310.}
To find a point equally distant from
the four vertices of a tetrahedron.
\end{Problem}

% [** Illustrations]
\begin{center}
\Input{123ab}{}
\end{center}

\Objective{Given the tetrahedron $P-ABC$.}
{To find a point $O$ equidistant from $P$, $A$, $B$, $C$.}

\begin{Construction}
At $D$, the middle point of $BC$, construct a plane
perpendicular to~$BC$.

This plane contains $E$, the center of the circle circumscribed
about $\triangle PBC$ and also~$F$, the center of the circle circumscribed
about $\triangle ABC$. Why?

In the plane $FDE$ draw $EG\perp ED$ and $FH\perp FD$.

Then $EG$ and $FH$ cannot be parallel and hence meet in some
point~$O$. Then~$O$ is the point required.
\end{Construction}

\Proof: Since $BC$ is $\perp$ to the plane $FDE$, it follows that each
of the planes $PBC$ and~$ABC$ is $\perp$ to plane $FDE$.\hfill §~\Sref{117}

Hence $OE$ is $\perp$ plane $PBC$ and $OF$ is $\perp$ plane $ABC$.\hfill §~\Sref{114}

Then $O$ is equidistant from $P$, $B$, and $C$, and also from $A$, $B$,
and~$C$. \\ % [** PP: Width-dependent line break]
\null\hfill §~\Sref{81}

Hence $OA = OB = OC = OP$.
\QEF

\begin{Corollary}{311.}{1.}
A sphere may be passed through any four
points, not all of which lie in the same plane.
\end{Corollary}

\begin{Corollary}{312.}{2.}
\index{Circumscribed@\textbf{Circumscribed}!sphere}%{\Pnum{312}}%
\index{Sphere@\textbf{Sphere}!circumscribed}%{\Pnum{312}}%
\index{Tetrahedron@\textbf{Tetrahedron}!inscribed}%{\Pnum{312}}%
A sphere may be circumscribed about any
tetrahedron.
\end{Corollary}

\NewFolio %% -----File: 124.png---Folio 116-------

\Section{TO FIND THE DIAMETER OF A SPHERE}

\begin{Problem}{313.}
To find the diameter of a given
material sphere.
\end{Problem}

With any point $P$ of the sphere as a pole, construct a
circle, and on this circle select any three points $A$, $B$, $C$.

Using a pair of compasses, measure the straight line-segments
$AB$, $BC$, $CA$, and construct the triangle $A'B'C'$ equal
to $ABC$.

Let $B'D'$ be the radius of the circle circumscribed about
$A'B'C'$.

If $PP'$ is the axis of the circle $ABC$ on the sphere and $BD$
the radius of this circle, then $BD = B'D'$.

% [** Illustrations]
\begin{center}
\Input{124ab}{}
\end{center}

Measure $PB$ by means of the compasses.

Then $PBP'$ is a right triangle, with $BD$ perpendicular to
its hypotenuse $PP'$.

$PB$ and $BD$ being known, we may now \emph{compute} $PD$ from
the right triangle $PBD$, and then compute $PP'$ from the similar
triangles $PBD$ and $PP'B$, for the latter using the relation
$PD : PB = PB : PP'$ or $PD × PP' = \overline{PB}^2$.

The segment $PP'$ may also be found by \emph{geometric} construction;
namely, by drawing a triangle equal to $P'BP$.

Show how to do this when $BP$ and $BD$ are known.

\NewFolio %% -----File: 125.png---Folio 117-------

\Section{INTERSECTING SPHERICAL SURFACES}

\begin{Theorem}{314.}{VI.}
The intersection of two spherical
surfaces is a circle.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{125ab}{}
\end{center}

\Proof: Two intersecting spherical surfaces may be developed
by two intersecting circles rotating about the line connecting
their centers $C$ and~$C'$. Let $A$ and $B$ be the two points common
to the two circles.

Then $BA$ is $\perp CC'$. \\
\null\hfill (\textit{$CC'$ is the $\perp$ bisector of $BA$}.)\hfill §~\Sref{28}

As the figure rotates about the line $CC'$, $AB$ remains $\perp CC'$
and therefore all its positions lie in a plane.\hfill §~\Sref{79}

Also $DB$ and $DA$ remain fixed in length.

Therefore the points $A$ and $B$ trace out a circle.
\QED

\begin{Sightwork}
\index{Loci@\textbf{Loci problems}|(}%

\item[1.] Find the locus of the centers of all spheres tangent to a given plane
at a given point.

\item[2.] Find the locus of the centers of all spheres of given radius tangent
to a given line at a given point.

\item[3.] Find the locus of the centers of all spheres of given radius tangent
to a fixed plane.

\item[4.] Find the locus of the centers of all spheres of given radius tangent
to a fixed line.

\item[5.] Find the locus of the centers of all spheres tangent to two given
intersecting planes.
\index{Loci@\textbf{Loci problems}|)}%

\end{Sightwork}

\NewFolio %% -----File: 126.png---Folio 118-------

\begin{Exercises}

\item[1.] The four lines perpendicular to the faces of a tetrahedron
at their circumcenters meet in a point.

\item[2.] The six planes perpendicular to the edges of a tetrahedron
at their middle points all meet in a point.

\item[3.] The planes bisecting the six dihedral angles of a tetrahedron
all meet in a point.

\item[4.] Show that a sphere may be inscribed in a cube.

\item[5.] Show that a sphere may be circumscribed about a cube.

\item[6.] Can a sphere be circumscribed about a rectangular
parallelopiped which is not a cube? Can a sphere be inscribed
in it? Prove.

\item[7.] If a plane $M$ is tangent to a sphere at a point $A$, show
that the plane of every great circle of the sphere through $A$
is perpendicular to $M$.

\item[8.] Show that the line of centers of two intersecting spheres
meets the spherical surfaces in the poles of their common circle.

\item[9.] Show that two spheres are tangent if they meet on their
line of centers. Distinguish two cases. State and prove the
converse of this proposition.

\item[10.] If a sphere is tangent to a given plane $M$ at a given
point $A$, how many additional points on the sphere are required
to determine it?

\begin{Suggestion}
Suppose one point $P$ given. Pass a plane through $P\perp$
to plane $M$ at~$A$. Is there only one such plane? Discuss fully.
\end{Suggestion}

\item[11.] Describe the set of all lines in space whose distances
from the center of a sphere are all equal to the radius of the
sphere.

\item[12.] Describe the set of all planes whose distances from the
center of a sphere are all equal to the radius of the sphere.

\item[13.] Describe the set of all spheres of given radius which
are tangent to a given sphere of greater radius.
\end{Exercises}

\NewFolio %% -----File: 127.png---Folio 119-------

\Section{SPHERICAL ANGLES}
\SecToc{Spherical Angles and Triangles}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{315.}{Spherical Angles.} Two planes through the center of a
\index{Angle@\textbf{Angle}!between two curves}%{\Pnum{315}}%
\index{Angle@\textbf{Angle}!spherical}%{\Pnum{315}}%
\index{Curves@\textbf{Curves}!angle between}%{\Pnum{315}}%
\index{Spherical@\textbf{Spherical}!angle}%{\Pnum{315}}%
sphere cut the surface of the sphere in two great circles which
intersect in two points and form four
\emph{spherical angles} about each of these
points. Any two of these angles are
\emph{adjacent} or \emph{vertical} as in the case of
angles formed by straight lines.
\end{minipage}%
% [** Illustration]
\hfill\Input{127a}{1.5in}

A spherical angle is measured by the
angle between the tangents to its sides
(arcs) at their common point.

\begin{Remark}
Only angles formed by great circles are considered in this book.
\end{Remark}


\Section{MEASURING A SPHERICAL ANGLE}

% [** Illustrations]
\begin{center}
\Input{127bc}{}
\end{center}

\begin{Theorem}{316.}{VII.}
A spherical angle is measured
by an arc of the great circle whose pole is the vertex
of the angle and which is intercepted by the sides of
the angle.
\end{Theorem}

\begin{Suggestion}
Show that $\wideparen{AB}$ measures the dihedral angle formed by
the planes $PAC$ and $PBC$ and that $\angle BCA = \angle TPR$.
\end{Suggestion}

\begin{Corollary}{317.}{1.}
The sum of the consecutive spherical
angles about a point is four right angles.
\end{Corollary}

\begin{Corollary}{318.}{2.}
A spherical angle is equal to the dihedral
angle formed by the planes of its arcs.
\end{Corollary}

\NewFolio %% -----File: 128.png---Folio 120-------

\Section{SPHERICAL POLYGONS}

% [****** PP: Width-dependent line break, set for 5in]
\Paragraph{319.}{Polygons and Triangles on a Sphere.} The portion of a
\index{Polygons@\textbf{Polygons}!spherical}%{\Pnum{319}}%
\index{Spherical@\textbf{Spherical}!polygon}%{\Pnum{319}}%
\index{Triangles@\textbf{Triangles}!symmetrical}%{\Pnum{319}}%
\begin{wrapfigure}[6]{r}{1.5in}
% [** Illustration]
\hfill\smash{\raisebox{-1in}{\Input{128a}{1.5in}}}
\end{wrapfigure}
spherical surface contained within a
polyhedral angle whose vertex is at the
center of a sphere is called a \emph{spherical
polygon}.

It follows that for every spherical
polygon there is a corresponding polyhedral
angle at the center of the sphere
made by drawing radii to the vertices
of the polygon. The face angles of the polyhedral angle correspond
to the sides of the spherical polygon and the dihedral
angles of the polyhedral angle to the angles of the spherical
polygon.

Since a plane through the center of a sphere intersects the
surface % [****** PP: Width-dependent line break, set for 5in]
\begin{wrapfigure}[7]{r}{1.5in}
% [** Illustration]
\hfill\smash{\raisebox{-1.125in}{\Input{128b}{1.5in}}}
\end{wrapfigure}
in a great circle, it follows that the sides of a spherical
polygon are \emph{arcs of great circles}.

\Indent Since a plane may be passed through
the vertex of a polyhedral angle such
that the polyhedral angle lies entirely
on one side of it, it follows that a
spherical polygon \emph{lies within one hemisphere}.

A spherical polygon of three sides is a \emph{spherical triangle}.

\Paragraph{320.}{Relation between the Parts of a Spherical Polygon and
the Corresponding Polyhedral Angle.}
\index{Corresponding@\textbf{Corresponding}!polyhedral angles}%{\Pnum{320}}%

(1)~The face angles of the polyhedral angle are measured by
the arcs forming the sides of the spherical polygon. Why?

(2)~The dihedral angles of the polyhedral angle are equal
in measure to the angles of the spherical polygon.\hfill §~\Sref{318}

\begin{Sightwork}

\item[] Why is the vertex of a polyhedral angle which is used in defining a
spherical polygon placed at the center of the sphere?
\end{Sightwork}

\NewFolio %% -----File: 129.png---Folio 121-------

\Section{COROLLARIES ON SPHERICAL POLYGONS}

The following propositions are now obvious corollaries of
the preceding definitions and discussions.

\Paragraph{321.}{} \textit{The sum of two sides of a spherical triangle is greater
than the third side}.

\begin{Remark}
This is a direct consequence of §~\Sref{151}.
\end{Remark}

\Paragraph{322.}{} \textit{The sum of the sides of a spherical polygon is less than
$360$°}.

\begin{Remark}
This is a direct consequence of §~\Sref{152}.
\end{Remark}

\Paragraph{323.}{} \textit{Each side of a spherical polygon is less than $180$°}.

\begin{Sightwork}

\item[1.] What is the angle between a meridian on the earth's surface and
the equator?

\item[2.] If two meridians are drawn meeting the equator $10$° apart, what
is the angle between these meridians?

\item[3.] If two meridians meet the equator $90$° apart, what can be said of
the three angles of the spherical triangle thus formed?

\item[4.] If in two spherical triangles the three sides of one are equal respectively
to the three sides of the other, what can be said of the face
angles of the corresponding trihedral angles?

\item[5.] If in two spherical triangles the three angles of one are equal respectively
to the three angles of the other, what parts are equal in the
corresponding trihedral angles?

\item[6.] If in two spherical polygons the angles of one are equal respectively
to the angles of the other, what parts are equal in the corresponding
polyhedral angles?

\item[7.] If in two spherical polygons the sides of one are equal respectively
to the sides of the other, what parts of the corresponding polyhedral
angles are equal?

\item[8.] Prove that if the sides of two spherical triangles are equal, then
the angles of the triangles are equal.

\begin{Suggestion} Use §~\Sref{144}.\end{Suggestion}
\end{Sightwork}

\NewFolio %% -----File: 130.png---Folio 122-------

\Section{SHORTEST SPHERICAL DISTANCE BETWEEN TWO POINTS}

\begin{Theorem}{324.}{VIII.}
\index{Distance@\textbf{Distance}!between two points on a sphere}%{\Pnum{324}}%
The shortest distance on a
sphere between two of its points is measured along
the minor arc of a great circle passing through these
points.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Proof: Let $A$ and $B$ be any two points
on a sphere, $AB$ the minor arc of a great
circle through them, and $ADCB$ any other
curve on the sphere connecting $A$ and $B$.
Let $D$ and~$C$ be any two points on the
curve $ADCB$ taken in order from $A$ to $B$.
\end{minipage}%
% [** Illustration]
\hfil\Input{130a}{1.25in}\hfil

Draw the great circle arcs $AD$, $AC$, $DC$, and~$CB$. Then
by §~\Sref{321} $\wideparen{AC} + \wideparen{CB} > \wideparen{AB}$ and $\wideparen{AD} + \wideparen{DC} > \wideparen{AC}$.

Hence, $\wideparen{AD} + \wideparen{DC} + \wideparen{CB} > \wideparen{AB}$.

If in like manner we subdivide $AD$, $DC$, $CB$, and continue
this process, we obtain a succession of paths, each longer than
the preceding, and thus we get closer and closer to the length
of the curve $ADCB$.

Hence, the curve $ADCB$ must be greater than $\wideparen{AB}$.
\QED

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{325.}{Symmetrical Spherical Triangles.}
\index{Polygons@\textbf{Polygons}!symmetrical}%{\Pnum{325}}%
\index{Symmetrical@\textbf{Symmetrical}!spherical triangles}%{\Pnum{325}}%
Two spherical triangles are \emph{symmetrical}
if the sides and angles of one are equal
respectively to the sides and angles of
the other, but arranged in the \emph{opposite}
order. Compare §~\Sref{146}.
\end{minipage}%
% [** Illustration]
\hfil\raisebox{-0.125in}{\Input{130b}{1.25in}\hfill}

\begin{Sightwork}

\item[1.] Is it possible to move the spherical triangle $ABC$ in the above
figure so as to make it coincide with triangle $A'B'C'$?

\item[2.] If in two plane triangles $ABC$ and $A'B'C'$ the corresponding parts
are equal, is it always possible to make the triangles coincide? Is this
always possible without inverting one of the triangles? Discuss the difference
in this respect between plane and spherical triangles.
\end{Sightwork}

\NewFolio %% -----File: 131.png---Folio 123-------

\Section{TRIHEDRAL ANGLES AND SPHERICAL TRIANGLES}

\begin{Theorem}{326.}{IX.}
Two equal or symmetrical trihedral
angles with vertices at the center of a sphere
intercept equal or symmetrical triangles, respectively,
on the sphere.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{131ab}{}
\end{center}
\vspace*{-0.25in}

\begin{Suggestion}[s for Proof]
In case the trihedral angles are equal they
can be made to coincide, whereby the spherical triangles will also be made
to coincide. The triangles are therefore equal. In case the trihedral
angles are symmetrical it follows directly from §§~\Sref{146}, \Sref{325} that the spherical
triangles are symmetrical.
\end{Suggestion}

\begin{Theorem}{327.}{X.}
If the radii drawn from the
vertices of a spherical triangle are extended, they meet
the sphere in the vertices of a triangle symmetrical to
the given triangle.
\end{Theorem}

The proof is left for the student.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Theorem}{328.}{XI.}
Two triangles
on the same sphere, or on equal
spheres, are equal or symmetrical
if three sides of one are equal respectively
to three sides of the other.
\end{Theorem}
\end{minipage}%
% [** Illustration]
\hfil\smash{\Input{131c}{1.25in}}\hfil

This is a direct corollary of §§~\Sref{145}, \Sref{149}. See the figure
under §~\Sref{326}.

\NewFolio %% -----File: 132.png---Folio 124-------

\Section{TRIANGLES EQUAL OR SYMMETRICAL}

\begin{Theorem}{329.}{XII.}
Two triangles on the same
sphere, or on equal spheres, are equal or symmetrical
if two sides and the included angle of one are equal
respectively to two sides and the included angle of the
other.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{132abc}{}
\end{center}

\Proof: This is a direct corollary of the theorems §§~\Sref{142}, \Sref{150}.

\Paragraph{330.}{Vertical Spherical Angles.} Spherical angles are \emph{vertical}
\index{Triangles@\textbf{Triangles}!vertical}%{\Pnum{319}}%
if they have the same vertex and if the sides of one are extensions
of the sides of the other.

\Paragraph{331.}{Right Spherical Triangle.} A \emph{right} spherical triangle
has one of its angles a right angle.
\index{Triangles@\textbf{Triangles}!right}%{\Pnum{331}}%

\Paragraph{332.}{Isosceles Spherical Triangle.} An \emph{isosceles} spherical
triangle has two equal sides.
\index{Isosceles@\textbf{Isosceles spherical triangle}}%{\Pnum{322}}%
\index{Triangles@\textbf{Triangles}!isosceles}%{\Pnum{322}}%

\begin{Sightwork}

\item[1.] Show that vertical spherical angles are equal.

\item[2.] What kind of spherical triangle is formed by two meridians on the
earth's surface and the arc of the equator which they intercept?

\item[3.] Show that every spherical triangle formed as described in Example~2
contains more than two right angles.

\item[4.] Show how to form a spherical triangle as in Example~2 in which
each angle shall be a right angle.

\item[5.] Show how to form a spherical triangle as in Example~2 in which
one angle shall be~179° and each of the other two~90°.
\end{Sightwork}

\NewFolio %% -----File: 133.png---Folio 125-------

\Section{EQUAL ANGLES OPPOSITE EQUAL SIDES}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Theorem}{333.}{XIII.}
The angles
opposite the equal sides of an isosceles
spherical triangle are equal.
\end{Theorem}

\Indent\begin{Suggestion}
Let $AC$ and $BC$ be the equal
sides. Draw $\wideparen{CD}$ to the middle point of $\wideparen{AB}$.
Then use §~\Sref{328}.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.125in}{\Input{133a}{1.25in}}

\begin{Corollary}{334.}{}
If two isosceles
spherical triangles are symmetrical, they are equal, and conversely.
\end{Corollary}

\begin{Exercises}

\item[1.] Compare fully the theorems on the equality of plane
triangles and of trihedral angles. Is there any theorem in
either case for which there is no corresponding theorem in
the other?

\item[2.] Compare in the same manner the theorems on the equality
of plane triangles and of spherical triangles.

\item[3.] Compare in the same manner the theorems on the equality
of trihedral angles and of spherical triangles.

\item[4.] If two face angles of a trihedral angle are equal, the
opposite dihedral angles are equal.

\item[5.] If two face angles of a trihedral angle are equal, it is
equal to its symmetrical trihedral angle.

\begin{Suggestion}
Compare the sides and then the angles of the corresponding
spherical triangles. Use §§~\Sref{333}, \Sref{329}.
\end{Suggestion}

\item[6.] Show how to find a pole of the circle through three
given points on a sphere.

\begin{Suggestion}
Let the given points be $A$, $B$, $C$. By §~\Sref{296} a pole of
the circle is equidistant from $A$, $B$, and~$C$. Connect $A$ and $B$ by an
arc of a great circle and construct another arc of a great circle bisecting
$\wideparen{AB}$ perpendicularly. Similarly construct a perpendicular bisector of
$\wideparen{BC}$. The points in which these two arcs meet will be the poles of the
circle through $A$, $B$, and~$C$.
\end{Suggestion}
\end{Exercises}

\NewFolio %% -----File: 134.png---Folio 126-------

\Section{POLAR TRIANGLES}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{335.}{Definition.} If with the vertices
\index{Polar@\textbf{Polar}!triangles}%{\Pnum{335}}%
\index{Triangles@\textbf{Triangles}!polar}%{\Pnum{335}}%
of a given spherical triangle as poles
arcs of great circles are constructed,
another spherical triangle is formed which
is called the \emph{polar triangle} of the first.

\Indent\begin{Remark}
Thus in the figure, $A$ is a pole of the arc
$B'C'$, $B$ is a pole of the arc $C'A'$, and $C$ is a
pole of the arc~$A'B'$. Hence $A'B'C'$ is the
polar triangle of triangle $ABC$.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfil\Input{134a}{1.5in}\hfil

\Paragraph{336.}{Polar Triangle. How Selected.} If with the vertices
$A$, $B$, $C$ of a spherical triangle as
poles three complete great circles
are constructed, each of these
circles meets each of the others
in two points, % [****** PP: Width-dependent line break, set for 5in]
\begin{wrapfigure}[11]{r}{2in}
% [** Illustration]
\Input{134b}{2in}
\end{wrapfigure}
thus forming eight
spherical triangles, as shown
in the figure, namely, $A'B'C'$,
$A'B'F$, $B'C'D$, $C'A'E$, $A'EF$,
$B'DF$, $C'DE$, and $DEF$.

There is one and only one of
these, namely, $A'B'C'$, such that
$A$ and~$A'$ are on the same side of circle $B'C'$, $B$ and~$B'$ on the
same side of circle $A'C'$, and $C$ and~$C'$ on the same side of
circle $A'B'$.

The triangle $A'B'C'$ as thus described is \emph{the polar triangle}
of $ABC$.

\begin{Sightwork}

\item[] In the above figure the parts of the great circles which are supposed
to be on the front side of the figure are given in solid lines while the
parts on the back side are dotted. Study the figure with care and state
which triangles are entirely on the front side, which are entirely on the
back side of the sphere, and which are partly on the front side and partly
on the back side of the sphere.
\end{Sightwork}

\NewFolio %% -----File: 135.png---Folio 127-------

\Section{TRIANGLES POLAR TO EACH OTHER}

\begin{Theorem}{337.}{XIV.}
If $A'B'C'$ is the polar triangle
of $ABC$, then $ABC$ is the polar triangle of
$A'B'C'$.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{135ab}{}
\end{center}
\vspace*{-0.25in}

\Objective{Given $\triangle A'B'C'$, the polar triangle of $ABC$.}
{To prove that $ABC$ is the polar triangle of $A'B'C'$.}

\Proof: (1)~$A'$ is at a quadrant's distance from $B$ because $B$ is
the pole of $\wideparen{A'C'}$. $A'$ is also at a quadrant's distance from~$C$
because $C$ is the pole of $\wideparen{A'B'}$.

Hence, $A'$ is the pole of $\wideparen{BC}$.\hfill §~\Sref{300}

Similarly, $B'$ is the pole of $\wideparen{AC}$ and $C'$ the pole of $\wideparen{AB}$.

(2)~To show that $A$ and~$A'$ lie on the same side of the circle
$BC$, we note that since $A$ is the pole of the circle $B'C'$ and
$A$ lies on the same side of this circle with $A'$, then $A$ and~$A'$
are at less than a quadrant's distance. Hence, it follows that
if $A'$ is at a quadrant's distance from $BC$, $A$ and~$A'$ must be
on the same side of~$BC$.

In like manner we show that $B$ and~$B'$ lie on the same side
of $AC$, and $C$ and~$C'$ on the same side of~$AB$.

\Paragraph{338.}{Corresponding Parts of Polar Triangles.} If $ABC$ and
\index{Corresponding@\textbf{Corresponding}!parts of polar triangles}%{\Pnum{338}}%
\index{Polar@\textbf{Polar}!triangles}%{\Pnum{338}}%
$A'B'C'$ are polar triangles, and if $A$ is a pole of $\wideparen{B'C'}$, then
$\angle A$ and $\wideparen{B'C'}$ are said to be \emph{corresponding parts}.

\NewFolio %% -----File: 136.png---Folio 128-------

\Section{MEASURE OF PARTS IN POLAR TRIANGLES}

\begin{Theorem}{339.}{XV.}
The sum of the measures of
an angle of a spherical triangle and the corresponding
arc of its polar triangle is~$180$°.
\end{Theorem}

% [** PP: Hard-coded formatting for Objectives environment]
\textbf{Given the polar triangles $ABC$ and $A'B'C'$. Denote the measures
in degrees of the angles by $A$, $B$, $C$, $A'$, $B'$, $C'$, and of the
corresponding sides by $a'$, $b'$, $c'$, $a$, $b$, $c$}.

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{136a}{}
\end{center}
\vspace*{-0.25in}

\begin{flalign*}
&\text{\itshape To prove that} &
  A + a' &= 180\text{°}\qquad   A' + a = 180\text{°} && \\
&&B + b' &= 180\text{°}\qquad   B' + b = 180\text{°} && \\
&&C + c' &= 180\text{°}\qquad   C' + c = 180\text{°} &&
\end{flalign*}

\textbf{Suggestions for Proof}: Extend (if necessary) arcs $A'B'$ and
$A'C'$ till they meet the great circle $BC$ in points $D$ and $E$,
respectively. Then arc $DE$ is the measure of $\angle A'$.

Also $\wideparen{BE} = 90$°, and $\wideparen{DC} = 90$°. (Why?)

But $\wideparen{BE} + \wideparen{DC} = \wideparen{BC} + \wideparen{ED} = a + A' = 180$°.

\begin{Corollary}{340.}{1.}
If in two spherical triangles an angle of
one is equal to an angle of the other, then the corresponding sides
of their polar triangles are equal; and conversely, if in two
spherical triangles a side of one is equal to a side of the other,
then the corresponding angles of their polar triangles are equal.
\end{Corollary}

\begin{Corollary}{341.}{2.}
If two spherical triangles are equal or
symmetrical, their polar triangles are equal or symmetrical.
\end{Corollary}

\NewFolio %% -----File: 137.png---Folio 129-------

\Section{SUM OF ANGLES OF A SPHERICAL TRIANGLE}

\begin{Theorem}{342.}{XVI.}
The sum of the angles of a
spherical triangle is less than six right angles and
greater than two right angles.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{137ab}{}
\end{center}

\Objective{Given the spherical triangle $ABC$.}
{To prove that \Inum{(1)}~$\angle A + \angle B + \angle C < 6$ rt.\ angles.\\
\indent\phantom{To prove that }\Inum{(2)}~$\angle A + \angle B + \angle C > 2$ rt.\ angles.}

\Proof: Construct the polar triangle $A'B'C'$, with sides
$a'$, $b'$, $c'$.

(1)~By §~\Sref{339} $\angle A + \angle B + \angle C + a' + b' + c' = 6 \text{ rt.\ } \Angles$.

Since $a' + b' + c' > 0$, $\therefore\quad \angle A + \angle B + \angle C < 6 \text{ rt.\ } \Angles$.

(2)~Using §~\Sref{322}, show that $\angle A + \angle B + \angle C > 2 \text{ rt.\ } \Angles$.

\begin{Corollary}{343.}{1.}
State and prove the theorem on trihedral
angles which corresponds to Theorem~XVI.
\end{Corollary}

\begin{Corollary}{344.}{2.}
The sum of the angles of a spherical
polygon of $n$~sides is greater than $2(n - 2)$ right angles and less
than $2n$ right angles.
\end{Corollary}

\Proof: Divide the polygon into $n - 2$ triangles. Then by
§~\Sref{342} the sum of the angles $> 2(n - 2) \text{ rt.\ } \Angles$.

Since each angle is less than two right angles, it follows
that the sum is less than $2n$ right angles.

\begin{Corollary}{345.}{3.}
State and prove the theorem on polyhedral
angles which corresponds to Corollary~2.
\end{Corollary}

\NewFolio %% -----File: 138.png---Folio 130-------

\Section{EQUAL AND SYMMETRICAL SPHERICAL TRIANGLES}

\begin{Theorem}{346.}{XVII.}
Two triangles on the same
sphere, or on equal spheres, are equal or symmetrical
if two angles and the included side of one are equal
respectively to two angles and the included side of the
other.
\end{Theorem}

\Proof. By §§~\Sref{340} and~\Sref{329} the polar triangles of the given
triangles are equal or symmetrical. Hence, by §~\Sref{341}, the
given triangles themselves are equal or symmetrical.

\begin{Corollary}{347.}{}
State and prove the theorem on trihedral
angles which corresponds to Theorem~XVII.
\end{Corollary}

\begin{Theorem}{348.}{XVIII.}
Two triangles on the same
sphere, or on equal spheres, are equal or symmetrical
if the angles of one are equal respectively to the angles
of the other.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{138ab}{}
\end{center}

\Proof: By §§~\Sref{340} and~\Sref{328} the polar triangles of the given
triangles are equal or symmetrical. Hence, by §~\Sref{341}, the
given triangles themselves are equal or symmetrical.

\begin{Corollary}{349.}{}
State and prove the theorem on trihedral
angles which corresponds to Theorem~XVIII.
\end{Corollary}

Is there a theorem on plane triangles corresponding to that
of §~\Sref{348}?

\NewFolio %% -----File: 139.png---Folio 131-------

\Section{CONSTRUCTION OF SPHERICAL TRIANGLES}

\begin{Problem}{350.}
On a given sphere to construct a
spherical triangle when its sides are given.
\end{Problem}

% [** Illustration]
\begin{center}
\Input{139a}{0.6\textwidth}
\end{center}

\textbf{Solution}. Let $O$ be the given sphere, and $a$, $b$, $c$ the sides of
the required triangle, and let $AA'$ be any diameter of the
sphere. With $A$ as a pole, construct circles $DBE$ and $FCG$,
whose polar distances from~$A$ are $c$ and~$b$ respectively.

With $B$ as a pole, construct a circle $HCK$, whose polar distance
from $B$ is~$a$. Then construct the three great circle arcs,
$AB$, $BC$, $CA$. $ABC$ is the required triangle. Why?

\begin{Sightwork}

\Page{129}%
\item[1.] What restrictions if any is it necessary to impose upon the three
given sides of the triangle in §~\Sref{350}? (See §§~\Sref{321}, \Sref{322}.)

\item[2.] In plane geometry two equal triangles may be constructed upon
the same base and on the same side of it. Is a corresponding construction
possible on the sphere?

\item[3.] If in the above construction each of two sides of the required triangle
is very great, that is, nearly a semicircle, show from the construction that
the third side must be very small.
\end{Sightwork}

\NewFolio %% -----File: 140.png---Folio 132-------

\Section{CONSTRUCTION OF SPHERICAL TRIANGLES}

\begin{Problem}{351.}
To construct a spherical triangle
when its three angles are given.
\end{Problem}

\textbf{Solution}. Let the three given angles be $A$, $B$, $C$, and let $a'$,
$b'$, $c'$ be arcs such that $a' + \angle A = 180$°, $b' + \angle B = 180$°, and $c' +
\angle C = 180$°. Then the triangle whose arcs are $a'$, $b'$, $c'$ will be
the polar triangle of the required triangle. This latter triangle
$A'B'C'$ may be constructed by the method of §~\Sref{350}. Then construct
the polar triangle of $A'B'C'$, which will be the required
triangle.

Give reasons in full for each step.

\begin{Problem}{352.}
To construct a trihedral angle when
its face angles are given.
\end{Problem}

\textbf{Solution}. Construct the corresponding spherical triangle by
the method of §~\Sref{350}.

Give the construction in full and prove each step.

\begin{Problem}{353.}
To construct a trihedral angle when
its dihedral angles are given.
\end{Problem}

\textbf{Solution}. Construct the corresponding spherical triangle by
the method of §~\Sref{351}.

Give reasons in full for each step.

\begin{Sightwork}

\item[1.] If two spherical triangles having angles respectively equal are constructed
on the same sphere, how are these triangles related? Prove.

\item[2.] If two trihedral angles with face angles respectively equal are constructed
as in §~\Sref{352}, how are they related? Prove.

\item[3.] If two trihedral angles with dihedral angles respectively equal are
constructed as in §~\Sref{353}, how are the trihedral angles related? Prove.

\item[4.] What restrictions if any must be placed upon the given angles
$A$, $B$, $C$ in §~\Sref{351}? Compare Example~1, page~\Pgref{129}.

\item[5.] What restrictions if any are needed in Examples 2 and~3?
\end{Sightwork}

\NewFolio %% -----File: 141.png---Folio 133-------

\begin{Exercises}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[1.] If two angles of a spherical triangle
are unequal, the sides opposite
them are unequal, the greater side being
opposite the greater angle.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{141a}{1.5in}}

\begin{Suggestion}
In the triangle $ABC$ let $\angle B$ be greater than $\angle A$. Draw
$\wideparen{BD}$, making $\angle ABD = \angle A$.
\end{Suggestion}

% [** PP: Not handling alignment]
Then, $\wideparen{AD} = \wideparen{BD}$, and $\wideparen{BD} + \wideparen{DC} > \wideparen{BC}$.

Hence, show that $\wideparen{AC} > \wideparen{BC}$.

\item[2.] State and prove a theorem on trihedral angles corresponding
to the preceding.

\item[3.] If the sides of a spherical triangle are $60$°, $80$°, $120$°, find
the angles of its polar triangle.

\item[4.] If the angles of a spherical triangle are $72$°, $104$°, $88$°,
find the sides of the polar triangle.

\item[5.] If a triangle is isosceles, prove that its polar triangle is
isosceles.

\item[6.] If each side of a spherical triangle is a quadrant, describe
its polar triangle.

\item[7.] Is it possible to construct a spherical triangle whose
sides are $50$°, $60$°, $120$°?

\item[8.] Is it possible to construct a spherical triangle whose
sides are $100$°, $120$°, $150$°?

\begin{Suggestion}
Consider the polar triangle of such triangle.
\end{Suggestion}

\item[9.] Consider the questions on trihedral angles corresponding
to the two preceding.

\item[10.] If the sides of a spherical triangle are $75$°, $95$°, and $115$°
respectively, find the angles of each triangle formed by the
polar construction.

\item[11.] If it is given that a spherical triangle is equilateral, can
we infer from the theorems thus far proved that its polar
triangle is equilateral?
\end{Exercises}

\NewFolio %% -----File: 142.png---Folio 134-------

\Section{SYMMETRICAL TRIANGLES ARE EQUAL IN AREA}

\Paragraph{354.}{Equal Areas Defined.} Two spherical polygons are said to
\index{Equal@\textbf{Equal}!areas}%{\Pnum{354}}%
\index{Triangles@\textbf{Triangles}!equal}%{\Pnum{354}}%
have equal areas if they can be made to coincide, or if they can
be divided into parts which can be made to coincide in pairs.

\begin{Remark}
Compare this with the definition of equal areas in Plane Geometry.
\end{Remark}

\begin{Theorem}{355.}{XIX.}
Two symmetrical spherical
triangles are equal in area.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{142a}{0.4\textwidth}
\end{center}

\Proof: Let $ABC$ be one of the given triangles. Extend the
radii $AO$, $BO$, $CO$ to meet the sphere in $A_1$, $B_1$, $C_1$, thus forming
a triangle symmetrical to $\triangle ABC$. \hfill §~\Sref{327}

Let $P$ be a pole of the circle through $A$, $B$, $C$. Extend $PO$
to meet the sphere in $P_1$. Draw $\wideparen{PA}$, $\wideparen{PB}$, $\wideparen{PC}$, and
$\wideparen{P_1A_1}$, $\wideparen{P_1B_1}$,
and $\wideparen{P_1C_1}$.

Suppose that $P$ lies within $\triangle ABC$.

The spherical triangles $PAB$, $PBC$, $PCA$, $P_1A_1B_1$, $P_1B_1C_1$,
$P_1C_1A_1$ are all isosceles.\hfill §~\Sref{296}

Now prove
\begin{gather*}
(1) \triangle PAB = \triangle P_1A_1B_1; \qquad (2) \triangle PBC = \triangle P_1B_1C_1; \\
(3) \triangle PCA = \triangle P_1C_1A_1. \\
\therefore\quad \text{area } \triangle ABC = \text{area } \triangle A_1B_1C_1.
\end{gather*}

But $\triangle A_1B_1C_1$ is equal to any other triangle which is symmetrical
to $\triangle ABC$.

$\therefore$ Two symmetrical spherical triangles are equal in area.

\NewFolio %% -----File: 143.png---Folio 135-------

\Section{BIRECTANGULAR SPHERICAL TRIANGLES}

\Paragraph{356.}{A birectangular spherical triangle} is one having two right
\index{Birectangular@\textbf{Birectangular spherical triangle}}%{\Pnum{356}}%
% [** PP: Index entry was Triangles, bi-rectangular; sole hyphenated instance]
\index{Triangles@\textbf{Triangles}!birectangular}%{\Pnum{356}}%
angles, as $\triangle PAB$ in the figure.

If the third angle of a birectangular triangle
is~$1$°, the triangle contains one of $720$ equal
parts of the surface of the sphere.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{357.}{Spherical Degree.} The area of a birectangular
\index{Degree@\textbf{Degree}!spherical}%{\Pnum{357}}%
\index{Spherical@\textbf{Spherical}!degree, minute, second}%{\Pnum{357}}%
triangle having a vertex angle of one
degree is called a \emph{spherical degree} and is used
as a unit of measure of areas of spherical polygons.

\Indent In a similar manner we define a \emph{spherical minute} and a
\emph{spherical second}.
\end{minipage}%
% [** Illustration]
\hfil\Input{143a}{1.25in}\hfil

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{358.}{The Lune.} A \emph{lune} is a figure formed
\index{Lune@\textbf{Lune}!angle of}%{\Pnum{358}}%
by two great semicircles having the same end-points.
The angle between these semicircles
is the \emph{angle of the lune}.

\Indent\begin{Corollary}{359.}{1.}
The area of a birectangular
spherical triangle in terms of spherical degrees
is equal to the number of degrees in the
third angle of the triangle.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfil\Input{143b}{1.25in}\hfil

\begin{Corollary}{360.}{2.}
The area of a lune in terms of spherical
degrees is twice the number of degrees in the angle of the lune.
\end{Corollary}

\Paragraph{361.}{Spherical Excess.} The number of degrees by which
the sum of the angles of a spherical triangle exceeds $180$° is
called the \emph{spherical excess} of the triangle.
\index{Spherical@\textbf{Spherical}!excess}%{\Pnum{361}}%

The spherical excess of a spherical polygon is the sum of its
angles less $(n - 2)180$°, where $n$ is the number of sides of the
polygon.

\Section{exercise}

\begin{Itemize}
\index{Problems@\textbf{Problems and Applications}}

\item[] Show that the spherical excess of a spherical polygon is less
than four right angles.

\end{Itemize}

\NewFolio %% -----File: 144.png---Folio 136-------

\Section{AREA OF A SPHERICAL TRIANGLE}
\SecToc{Area of the Sphere}

\begin{Theorem}{362.}{XX.}
\index{Area@\textbf{Area}!of a spherical triangle}%{\Pnum{362}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{362}}%
The area of a spherical triangle
in terms of spherical degrees is equal to its spherical
excess.\footnotemark
\end{Theorem}
\footnotetext{This theorem was discovered by Cavalieri. See \hyperref[page:Cavalieri]{Frontispiece}.}
\Page{134}%

% [** Illustration]
\begin{center}
\Input{144a}{0.4\textwidth}
\end{center}

\Proof: We are to show that area $\triangle ABC = \angle A + \angle B + \angle C
- 180$°.

By §~\Sref{360}, the areas of the lunes $ACDB$, $CAEB$, $BCFA$ in
spherical degrees are as follows:
\begin{align*}
\triangle ABC + \triangle BCD &= ACDB = 2 \angle A. \\
\triangle ABC + \triangle BAE &= CAEB = 2 \angle C. \\
\triangle ABC + \triangle CFA &= BCFA = 2 \angle B.
\end{align*}

Hence, adding,

$3\triangle ABC + \triangles BCD,\ BAE,\ CFA = 2(\angle A + \angle B + \angle C)$.

Now $\triangles BCD$ and $AEF$ are symmetrical and equal in area.

Hence, \\
$2 \triangle ABC + \triangles ABC,\ AEF,\ BAE,\ CFA = 2(\angle A + \angle B + \angle C)$.

But $\triangles ABC,\ AEF,\ BAE,\ CFA$ together constitute a hemisphere
or $360$~spherical degrees.

Hence, $2 \triangle ABC + 360\text{°} = 2 (\angle A + \angle B + \angle C)$.

Solving, $\triangle ABC = \angle A + \angle B + \angle C - 180$°.
\QED

\medskip
\noindent\hrule
\medskip

\textsc{Note}.---The spherical degree differs fundamentally from the units
of measure hitherto used. This unit is a certain fraction of the surface
of the sphere and hence its actual size depends upon the size of the
sphere.

\NewFolio %% -----File: 145.png---Folio 137-------

\Section{AREA OF A SPHERICAL POLYGON}

\begin{Theorem}{363.}{XXI.}
\index{Area@\textbf{Area}!of a spherical polygon}%{\Pnum{363}}%
The area of a spherical polygon
in terms of spherical degrees is equal to its spherical
excess.
\end{Theorem}

\Proof: Join one vertex of the polygon to each non-adjacent
vertex, thus forming $n - 2$ spherical triangles. Now since the
sum of the spherical excesses of these triangles is the spherical
excess of the polygon, the conclusion is evident.

\begin{Sightwork}

\item[1.] What is the area in spherical degrees of a birectangular triangle
if one of its angles is $54$°? if one angle is $24$°? if one angle is $36$°?

\item[2.] What is the spherical excess of a triangle the sum of whose angles
is $285$°?

\item[3.] What is the spherical excess of a triangle whose angles are $75$°,
$110$°, and $150$°?

\item[4.] What is the spherical excess of a spherical hexagon the sum of
whose angles is $1060$°?

\item[5.] What is the spherical excess of a spherical pentagon whose angles
are $90$°, $120$°, $75$°, $108$°, and $165$°?

\item[6.] What is the angle of a lune whose area is $160$~spherical degrees? %[** PP: Changed to question mark]

\item[7.] The spherical excess of a triangle is $120$°. Two of its angles are
$110$° and $108$° respectively. Find the third angle.

\item[8.] Between what limits is the sum of the angles of a spherical polygon
of eight sides?

\item[9.] If the sum of the angles of a spherical polygon is $11$ right angles,
what is known about the number of its sides?

\item[10.] If the sum of the angles of a spherical polygon is $14$ right angles,
what is known about the number of its sides?

\item[11.] The sides of a spherical triangle are $85$°, $95$°, $110$°. Find the
area of each of the eight triangles formed by the polar construction from
this triangle.

\item[12.] The area of a spherical triangle is $74$ spherical degrees. One
angle is $105$°. Of the other two angles one is twice the other. Find all
the angles of the triangle.
\end{Sightwork}

\NewFolio %% -----File: 146.png---Folio 138-------

\Section{LATERAL AREA OF A FRUSTUM OF A CONE}

\begin{Theorem}{364.}{XXII.}
The lateral area of a frustum
of a right circular cone is equal to the altitude of
the frustum multiplied by the circumference of a circle
whose radius is the perpendicular distance from a point
in the axis of the frustum to the middle point of an
element.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{146ab}{}
\end{center}

\Objective{Given a frustum with an element $AA'$ whose middle point is $B$,
$CC'$ the axis of the frustum, $BD \perp CC'$ and $EB \perp AA'$.}
{To prove that the lateral area is equal to $2\pi\cdot EB\cdot CC'$.}

\Proof: By §~\Sref{269}, the lateral area is $2\pi\cdot BD\cdot AA'$.

Hence, we must show that
\[
EB \cdot CC' = BD \cdot AA'.
\]

\noindent\begin{minipage}[b]{\textwidth-1.25in}
\Indent To do this, draw $A'F \perp AC$ and show that
$\triangle AFA' \sim \triangle EDB$.

\Indent\begin{Corollary}{365.}{}
The lateral area of a right
circular cone is equal to its altitude times the length
of a circle whose radius is the perpendicular from
a point in the axis to the middle point of an element.
\end{Corollary}
\end{minipage}%
% [** Illustration]
\hfil\Input{146c}{1in}\hfil

{\small\textit{Note}. The above theorem and corollary are needed in deducing the
area of the surface of a sphere. We have already computed the area of a
spherical triangle in terms of \emph{spherical degrees}, but we now wish to derive
the area of the spherical surface in terms of \emph{plane units of measure}.}

\NewFolio %% -----File: 147.png---Folio 139-------

\Section{CIRCUMSCRIBED AND INSCRIBED CONES AND FRUSTUMS}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{366.}{} About a circle circumscribe a polygon as follows: Construct
\index{Circumscribed@\textbf{Circumscribed}!cone}%{\Pnum{366}}%
\index{Generator@\textbf{Generator}!of a spherical surface}%{\Pnum{366}}%
\index{Inscribed@\textbf{Inscribed}!cone}%{\Pnum{366}}%
two diameters $AB$ and $CD$ at right angles to each other
and divide each quadrant into an \emph{even} number of parts by
points, as $E$, $F$, $G$. At each alternate
division point, beginning with the first
point~$E$, draw a tangent.

\Indent There results a regular polygon with
two vertices on each of the diameters
$AB$ and $CD$ extended.
\end{minipage}%
% [** Illustration]
\hfill\Input{147a}{1.5in}

If now we construct another polygon
in the same manner by dividing each
quadrant into \emph{twice} as many arcs, it will
likewise have two vertices on each of
the diameters $AB$ and $CD$ extended. The vertices on $AB$ will
lie between $M$ and~$N$.

This construction may be repeated at pleasure, thus obtaining
a sequence of polygons each lying closer to the circle than
the preceding.

Now inscribe a polygon similar to the first one of the set just
circumscribed, by joining the points $C$ and $F$, $F$ and $A$, and so on,
repeating this process to form a sequence of inscribed polygons.

If now the whole figure is made to revolve about $AB$ as an
axis, the circle generates a sphere, and the circumscribed and
inscribed polygons generate sets of circumscribed and inscribed
cones and frustums of cones.

\Paragraph{367.}{Fundamental Assumption on the Area of a Sphere.} We
assume that
\index{Axioms@\textbf{Axioms}}%{\Pnum{367}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{367}}%

\textit{A sphere has a definite area which is less than the surface of
any circumscribed figure and greater than the surface of any
inscribed convex figure}.

\begin{Remark}
The student should note that while the statement just preceding is
obviously true, it is not capable of proof by pure deduction. It is therefore
\emph{necessarily} in the nature of an \emph{assumption} or \emph{axiom}.
\end{Remark}

\NewFolio %% -----File: 148.png---Folio 140-------

\Section{FORMULA FOR THE AREA OF A SPHERE}

\begin{Theorem}{368.}{XXIII.}
\index{Area@\textbf{Area}!of a sphere}%{\Pnum{368}}%
\index{Sphere@\textbf{Sphere}!area of}%{\Pnum{368}}%
The area of a sphere whose
radius is $r$ is $4\pi r^2$.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\Proof: (a)~Denote by $r$ the radius of the circle which generates
the sphere. That is, in the figure, $r = OG = OE = OH$.

\Indent By §~\Sref{364} the lateral areas of the outside frustums whose axes
are $OL$ and $OK$ are $2\pi r\cdot OL$ and $2\pi r\cdot OK$, and by §~\Sref{365} the
lateral areas of the cones whose axes are $LN$ and $KM$ are respectively
$2\pi r\cdot LN$ and $2\pi r\cdot KM$.
\end{minipage}%
% [** Illustration]
\hfil\Input{148a}{1.5in}\hfil

Hence, the total surface of the whole circumscribed figure is
\begin{flalign*}
&&          &2\pi r\cdot MK + 2\pi r\cdot KO + 2\pi r\cdot OL + 2\pi r\cdot LN, && \\
&\text{or}& &2\pi r (MK + KO + OL + LN) = 2\pi r\cdot MN. &&
\end{flalign*}

If now a polygon of twice the number of sides is constructed,
as described in §~\Sref{366}, we obtain another circumscribed
figure whose area is $2\pi r\cdot M'N'$ where $M'$ and $N'$ are the two
vertices on the line $AB$ extended, but lying between $M$ and $N$,
so that $2\pi r\cdot M'N' < 2\pi r\cdot MN$.

As this process goes on, the total surface generated decreases
and may be made to approximate \emph{as nearly as we please} to
\[
2\pi r × AB = 2\pi r × 2r = 4\pi r^2.
\]
Therefore, the area of the sphere cannot be greater than $4\pi r^2$.

\NewFolio %% -----File: 149.png---Folio 141-------

(b)~Using the same figure, let $OG'$ be the apothem of the
inscribed polygon.

Then, as under~(a), we find that the area of the figure developed
by revolving the inscribed polygon about $AB$ is
\[
2\pi \cdot AB \cdot OG' = 4\pi r \cdot OG'.
\]
By continuing to double the number of sides, $OG'$ may be made
to approach as nearly as we please to $OG = r$, $OG'$ being
always less than $OG$.

Hence, the total surface developed increases and approaches
\emph{as nearly as we please to}
\[
4\pi r × OG = 4\pi r^2.
\]

Therefore the area of the sphere cannot be less than $4\pi r^2$.

Since from (a) and~(b) the area of the surface of the sphere
can be neither greater than $4\pi r^2$ nor less than $4\pi r^2$, it follows
that it is exactly equal to $4\pi r^2$.
\QED

\begin{Sightwork}

\item[] Use the value $\pi = 3.1416$. % [** PP: Remove comma after value]

\item[1.] Find the surface of a sphere whose radius is one inch.

\item[2.] Find the surface of a sphere whose radius is $10$~inches.

\item[3.] What is the relation between the surface of a sphere and the area
of a circle of the same radius?

\item[4.] Find the area in square inches of a spherical degree on a sphere of
radius $15$~inches.

\item[5.] What fraction of the surface of a sphere is occupied by a birectangular
triangle having one angle of~$65$°? by a lune with an angle of~$84$°?

\item[6.] Find the area in square inches of a birectangular spherical triangle
with one angle equal to~$35$°, if the triangle is on a sphere of radius
$10$~inches.

\item[7.] Find the area in square inches of a lune whose angle is~$42$°, if the
lune is on a sphere of radius $20$~inches.
\end{Sightwork}

\NewFolio %% -----File: 150.png---Folio 142-------

\Section{EXERCISES IN ARITHMETIC COMPUTATION}

\Paragraph{Example.}{} Find the area in square inches of a spherical
triangle whose angles are $80$°, $85$°, and~$97$° if the radius of the
sphere is $6$~inches.

\textit{Solution}.\quad The spherical excess of the triangle is
\[
80\text{°} + 85\text{°} + 97\text{°} - 180\text{°}
  = 262\text{°} - 180\text{°} = 82\text{°}.
\]
The total area of the sphere is $4\pi × 6^2 = 452.3904$~sq.~in., and one spherical
degree is $\tfrac{1}{720}$ of $452.3904 = .62832$~sq.~in.

Hence $82$ spherical degrees is $82 × .62832 = 51.52224$~sq.~in.

\begin{Itemize}

\item[1.] Find the area in square inches of a spherical triangle
whose angles are $70$°, $80$°, $90$° if the radius of the sphere is
$10$~inches.

\begin{Suggestion}
First find the spherical excess of the triangle.
\end{Suggestion}

\item[2.] Find the area in square inches of a spherical triangle
whose angles are $95$°, $110$°, $75$° if the radius of the sphere is
$15$~inches.

\item[3.] Find the area of a spherical polygon whose angles are
$110$°, $120$°, $130$°, and~$95$° if the radius of the sphere is $8$~inches.

\item[4.] Find the area of a spherical polygon whose angles are
$130$°, $140$°, $110$°, $100$°, $160$°, and~$150$° if the radius of the sphere
is $20$~inches.
\end{Itemize}


\Section{EXERCISES IN ALGEBRAIC COMPUTATION}

\begin{Itemize}

\item[1.] The area of a sphere is $s$~square inches. Find the radius
of the sphere in terms of~$s$.

\item[2.] The spherical excess of a spherical triangle is~$e$ and the
radius of the sphere is~$r$. Find the area of the triangle in
terms of $e$ and~$r$.

\item[3.] The spherical excess of a spherical polygon is $e$ and its
area is $a$~square inches. Find the radius of the sphere in
terms of $e$ and~$a$.

\item[4.] The area in square inches of a spherical triangle is $a$ and
the radius of the sphere is~$r$. Find the spherical excess in
terms of $a$ and~$r$.
\end{Itemize}

\NewFolio %% -----File: 151.png---Folio 143-------

\Section{THE VOLUME OF A SPHERE BY INSPECTION}
\SecToc{Volume of the Sphere}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\Paragraph{369.}{} The formula for the volume of a sphere may be inferred
directly from the accompanying figure.
\index{Sphere@\textbf{Sphere}!volume of}%{\Pnum{369}}%
\index{Volume@\textbf{Volume}!of a sphere}%{\Pnum{369}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{369}}%

The sphere is covered with a network
of spherical quadrilaterals. If these are
taken small enough, they may be regarded
as approximately \emph{plane surfaces}.
\end{minipage}%
% [** Illustration]
\hfil\Input{151a}{1.5in}\hfil

On this supposition we have a set of
pyramids with a common altitude~$r$ and
the sum of their bases approximately
equal to the area of the sphere.

Hence, their combined volume is $\frac{1}{3} r × (\text{area of sphere})$
or $\tfrac{1}{3} r \cdot 4\pi r^2$. That is, the volume is $\tfrac{4}{3} \pi r^3$.

It is clear that, by making these quadrilaterals sufficiently
small each one may be made to approach as nearly as we
please to a plane surface. Hence, each pyramidal figure with
vertex at $O$ is made to approach the form of a true pyramid
as nearly as we please.

\Paragraph{370.}{Fundamental Assumption on the Volume and Surface of a
Sphere.} In the formal proof on page~\Pgref{142}, we use polyhedrons
circumscribed about the sphere and we assume that
\index{Axioms@\textbf{Axioms}}%{\Pnum{370}}%

\textit{The surface and the volume of a sphere may be approximated
as nearly as we please by taking the surfaces and the volumes
of a series of circumscribed polyhedrons all of whose faces are
made to decrease indefinitely}.

\begin{Remark}
\textit{Note}. This assumption, when taken together with §~\Sref{367}, implies that
the surface obtained by taking the surfaces of a series of polyhedrons
is the same as the surface obtained by taking the surfaces of the circumscribed
figures described in §~\Sref{366}. This is of course obvious at a glance,
though a formal deductive proof is very difficult.
\end{Remark}

\Section{exercise}

\begin{Itemize}
\index{Problems@\textbf{Problems and Applications}}

\item[] Using §~\Sref{369} find the volume of a sphere whose radius is
$10$~in.; also one whose diameter is $6$~ft.

\end{Itemize}

\NewFolio %% -----File: 152.png---Folio 144-------

\Section{FORMULA FOR THE VOLUME OF A SPHERE}

\begin{Theorem}{371.}{XXIV.}
\index{Circumscribed@\textbf{Circumscribed}!polyhedron}%{\Pnum{371}}%
\index{Polyhedrons@\textbf{Polyhedrons}!circumscribed}%{\Pnum{371}}%
\index{Sphere@\textbf{Sphere}!volume of}%{\Pnum{371}}%
The volume of a sphere
whose radius is $r$ is $\tfrac{4}{3} \pi r^3$.
\end{Theorem}

\begin{center}
\Input{152ab}{}
\end{center}

\Page{142}%
\Proof: Consider a sphere of radius~$r$ with a cube circumscribed
about it. Connect its vertices with the center of the
% [**Illustration]
sphere. Then the cube is divided into six pyramids, each having
an altitude~$r$. Since the volume of each pyramid equals
the area of its base times one third its altitude, it follows that
the sum of the volumes of these pyramids is equal to the sum
of their bases times one third their common altitude.

Consider now a polyhedron obtained from this cube by cutting
off its eight vertices by planes tangent to the sphere.
This new polyhedron is circumscribed about the sphere and
approaches it more nearly than the cube. Continuing in this
manner\footnoteTrans{See \Tref{152a}{Transcriber's Notes}.}
\Erratum{err152a} % [** PP: Not totally clear what's meant, mathematically]
we may obtain a series of circumscribed polyhedrons
approaching the sphere as nearly as we please.

By joining the vertices of each polyhedron to the center of
the sphere we obtain a set of pyramids, each with altitude~$r$.

Suppose the total surfaces of these successive polyhedrons
are $s_1$, $s_2$, $s_3$, $s_4$, $\dotsc$. Then their volumes are $\tfrac{r}{3}s_1$, $\tfrac{r}{3}s_2$, $\tfrac{r}{3}s_3$, $\dotsc$.
But, by §~\Sref{370}, $s_1$, $s_2$, $s_3$, $\dotsc$ approach the surface of the sphere,
or $4\pi r^2$, as nearly as we please. Hence the successive volumes
approach $\tfrac{r}{3} × 4\pi r^2 = \tfrac{4}{3} \pi r^3$ as nearly as we please. That is,

% [** PP: Bold math in the original]
\textbf{Surface of a sphere ${} = 4\pi r^2$, and Volume of a sphere $ {} = \tfrac{4}{3} \pi r^3$}.

\NewFolio %% -----File: 153.png---Folio 145-------

\begin{Sightwork}

\item[1.] Given a sphere of radius $6$~inches, is there any upper limit to the
volume of its circumscribed polyhedrons? That is, can polyhedrons be
circumscribed having a volume as large as we please?

\item[2.] With the same sphere is there any lower limit to the volume of its
circumscribed polyhedrons?

\item[3.] Show that the areas of two spheres are in the same ratio as the
squares of their radii or of their diameters.

\item[4.] Show that the volumes of two spheres are in the same ratio as the
cubes of their radii or of their diameters.

\item[5.] Express the area of a sphere whose radius is $8$~in.\ in terms of~$\pi$.

\item[6.] Express the volume of a sphere whose radius is $10$~ft.\ in terms
of~$\pi$.

\item[7.] The surface of a polyhedron circumscribed about a sphere of
radius $4$~in.\ is $420$~sq.~in. Find its volume.

\end{Sightwork}

\Section{EXERCISES IN ARITHMETIC COMPUTATION}

\begin{Itemize}

\item[1.] The volume of a polyhedron circumscribed about a
sphere of radius $3.5$~in.\ is $450$~cu.~in. Find its surface.

\item[2.] If the area of a sphere is $227$~sq.~ft., find its radius.

\item[3.] If the volume of a sphere is $335$~cu.~in., find its radius.

\item[4.] If the volumes of two spheres are $27$~cu.~in.\ and $729$~cu.~in.,
find the ratio of their radii.

\end{Itemize}


\Section{EXERCISES IN ALGEBRAIC COMPUTATION}

\begin{Itemize}

\item[1.] The volume of a sphere is $v$~cubic inches. Find its
radius in terms of $v$ and~$\pi$.

\item[2.] The area of the surface of a sphere is s square inches.
Find its volume in terms of $s$ and~$\pi$.

\item[3.] The area of the surface of a sphere is numerically equal
to its volume. Find the radius of the sphere if an inch is
the unit of measure.

\item[4.] The difference between the volume of a cube and that
of its inscribed sphere is~$v$. Find the radius of the sphere in
terms of $v$ and~$\pi$.

\end{Itemize}

\NewFolio %% -----File: 154.png---Folio 146-------

\Section{SPHERICAL SEGMENTS, CONES, SECTORS}

\Paragraph{372.}{Zone, Segment.} That part of a spherical surface included
\index{Altitude@\textbf{Altitude}!of a zone}%{\Pnum{372}}%
\index{Segment@\textbf{Segment}!spherical}%{\Pnum{372}}%
\index{Spherical@\textbf{Spherical}!segment}%{\Pnum{372}}%
\index{Zone@\textbf{Zone}!altitude of}%{\Pnum{372}}%
\index{Zone@\textbf{Zone}!bases of}%{\Pnum{372}}%
\index{Zone@\textbf{Zone}!of one base}%{\Pnum{372}}%
between two parallel planes cutting it is called a
\emph{zone}. The perpendicular distance between the planes is the
\emph{altitude} of the zone and of the corresponding segment.

The portion of a sphere included between two parallel
planes cutting it is called a \emph{spherical segment}, and the two
circular sections made by the parallel planes are its \emph{bases}.
\index{Altitude@\textbf{Altitude}!of a spherical segment}%{\Pnum{372}}%
\index{Base@\textbf{Base}!of a spherical segment}%{\Pnum{372}}%

If one of the cutting planes is tangent to the sphere, then
the spherical segment and the corresponding zone are said to
have but \emph{one base}. The altitude in this case is the perpendicular
distance from the base to the tangent plane.

\noindent\begin{minipage}[b]{\textwidth-1.375in}
\Indent\Paragraph{373.}{Spherical Cone, Sector.} If one nappe
\index{Base@\textbf{Base}!of a spherical sector}%{\Pnum{373}}%
\index{Cone@\textbf{Cone}!spherical}%{\Pnum{373}}%
\index{Sector@\textbf{Sector}!spherical}%{\Pnum{373}}%
\index{Spherical@\textbf{Spherical}!cone}%{\Pnum{373}}%
\index{Spherical@\textbf{Spherical}!sector}%{\Pnum{373}}%
of a convex conical surface has its vertex
at the center of a sphere, the portion of
the sphere cut out by this surface is called
a \emph{spherical cone}.
% [** Illustration]

\Indent If two spherical cones have the same axis,
one lying within the other, the figure formed
by their two lateral surfaces, together with
the part of the sphere intercepted between
them, is called a \emph{spherical sector}.

If the two cones are right circular cones, they
intercept circles on the sphere, and the zone thus
included is called the \emph{base} of the spherical sector.
\end{minipage}%
% [** Illustration]
\hfill\Input{154ab}{1.25in}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.125in}
\Indent\begin{Remark}
If the accompanying figure be revolved about $LM$ as
an axis, then any arc, as $GD$ or $MD$, generates a zone,
the former with two bases, the latter with one.

The figure $MDF$ of $FDGE$ generates a spherical segment,
the former with one base, the latter with two.

The figure $CAL$ or $CBL$ generates a spherical cone,
$CL$ being the common axis.

The figure $CBA$ generates a spherical sector and arc $AB$ generates the
zone which is the base of the spherical sector.
\end{Remark}
\end{minipage}%
% [** Illustration] [** PP: M label illegible]
\hfill\Input{154c}{1in}

\NewFolio %% -----File: 155.png---Folio 147-------

\Section{AREA OF A ZONE. VOLUME OF A SPHERICAL CONE}

\Paragraph{374.}{Area of a Zone.} An argument precisely like that of
\index{Area@\textbf{Area}!of a zone}%{\Pnum{374}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{374}}%
§~\Sref{368} shows that the area of a zone is
\[
s = 2\pi rh,
\]
where $h$ is the altitude of the zone.

That is, instead of $AB$, the diameter in case of the sphere,
we should have the sum of the altitudes of the frustums circumscribed
about the zone equal to~$h$, the altitude of the zone.

\Paragraph{375.}{Volume of a Spherical Cone and a Spherical Sector.} An
\index{Volume@\textbf{Volume}!of a spherical cone}%{\Pnum{375}}%
\index{Volume@\textbf{Volume}!of a spherical sector}%{\Pnum{375}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{375}}%
argument precisely like that of §~\Sref{371} shows that the volume
of a spherical cone is
\[
v = \dfrac{r}{3} \cdot s,
\]
where $s$ is the area of the zone of one base which is cut out of
the sphere by the cone. Hence, if $h$ is the altitude of this
zone, we have
\[
v = \dfrac{r}{3} \cdot 2\pi rh = \dfrac{2\pi}{3} r^2h.
\]

In like manner the \emph{volume of a spherical sector} is
\[
v = \dfrac{2\pi}{3} r^2h,
\]
where $h$ is the altitude of the zone of two bases which is cut
out by the sector.


\begin{Exercises}

\item[1.] The radius of a sphere is $6$~in.\ and the altitude of a zone
is $5$~in. Find the area of the sphere and of the zone.

\item[2.] The area of a zone is $36\pi$~sq.~ft.\ and its altitude $4$~ft.
Find the radius of the sphere.

\item[3.] On a sphere of radius $8$~in.\ a spherical cone cuts out a
zone whose altitude is $2$~in. Find the volume of the cone.

\item[4.] Find the volume of the spherical sector cut out of a
sphere of radius $9$~in., if the altitude of the zone is $2$~in.
\end{Exercises}

\NewFolio %% -----File: 156.png---Folio 148-------

\Section{VOLUME OF A SPHERICAL SEGMENT}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\begin{Problem}{376.}
\index{Volume@\textbf{Volume}!of a spherical segment}%{\Pnum{376}}%
To find the volume of a spherical
segment.
\end{Problem}

\Indent\textbf{Solution}. Let $r$ be the radius of the
sphere, and $r_1$ and $r_2$ the radii of the
bases of the segment, $h$ the altitude of
the segment, and let the segment be generated
by revolving the figure $ACDB$ about
$AO$ as an axis.
\end{minipage}%
% [** Illustration]
\hfil\Input{156a}{1.5in}\hfil

\begin{flalign*}
&\indent\text{We have Vol.\ generated by } & ODB &= \dfrac{2\pi}{3} r^2h. && && \tag*{§~\Sref{375}} \\
&\indent\text{\phantom{We have }Vol.\ generated by\ } & OAB &= \dfrac{\pi}{3} r_1^2(h+d). && && \tag*{(Why?)} \\
&\indent\text{\phantom{We have }Vol.\ generated by\ } & OCD &= \dfrac{\pi}{3} r_2^2d. && && \tag*{(Why?)} \\
\end{flalign*}
\begin{flalign*}
&\indent\text{Hence,} & v &= \dfrac{2\pi}{3} r^2h + \dfrac{\pi}{3} r_1^2(h+d) - \dfrac{\pi}{3} r_2^2d && \\
&&                        &= \dfrac{\pi}{3} \bigl[2r^2h + r_1^2h + d(r_1^2 - r_2^2)\bigr]. && \tag*{(1)} \\
&\indent\text{From} & r^2 &= r_2^2 + d^2 \text{ and } r^2 = r_1^2 + (h+d)^2 && \\
&\text{we obtain} & d &= \dfrac{r_2^2 - r_1^2 - h^2}{2h}. && \tag*{(2)}
\end{flalign*}

Substituting this value of $d$ in $r^2 = r_2^2 + d^2$, \\
we get\hfill $r^2 = \dfrac{r_2^4 + r_1^4 + h^4 - 2r_1^2r_2^2 + 2h^2r_2^2 + 2h^2r_1^2}{4h^2}$.\hfill (3)

Substituting (2) and~(3) in~(1) and reducing, we have
\[
v = \dfrac{\pi h}{2} (r_1^2 + r_2^2) + \dfrac{\pi}{6} h^3. \tag*{\QEF}
\]

\begin{Corollary}{377.}{}
The volume of a spherical segment of one
base is
\[
v = \pi h^2 \left( r - \dfrac{h}{3} \right).
\]
\end{Corollary}

\begin{Suggestion}
In this case $r_1^2 = 0$ and $r_2^2 = r^2 - (r - h)^2 = 2rh - h^2$.
Substituting in the formula above, we have the result.
\end{Suggestion}

\NewFolio %% -----File: 157.png---Folio 149-------

\Section{SUMMARY OF BOOK V}
\index{Summaries@\textbf{Summaries}}

\begin{Itemize}

\item[1.] Define sphere, diameter, radius.

\item[2.] Collect the theorems of Book~V involving plane sections
of the sphere.

\item[3.] Define axis and pole of a circle, great circle, polar distance.

\item[4.] Define tangent plane to a sphere, inscribed and circumscribed
polyhedrons.

\item[5.] Arrange in parallel columns the corresponding theorems
on trihedral angles and spherical triangles which are proved
without the use of polar triangles.

\item[6.] Collect the definitions on polar triangles.

\item[7.] Collect the theorems on polar triangles.

\item[8.] Continue the lists begun in Example~5, adding the
theorems proved by means of polar triangles.

\item[9.] Make a list of the definitions involving polyhedral angles
and spherical polygons.

\item[10.] Collect the theorems involving polyhedral angles and
spherical polygons.

\item[11.] Collect the theorems on the areas of spherical triangles
and polygons.

\item[12.] Give the definitions and assumptions pertaining to the
area and volume of the sphere.

\item[13.] State all the theorems pertaining to the area and volume
of the sphere.

\item[14.] Give the definitions and theorems pertaining to spherical
figures, such as zones, cones, sectors, segments.

\item[15.] Collect all the mensuration formulas in this Book.

\item[16.] Collect all the mensuration formulas of solid geometry.

\item[17.] Describe some of the most important applications in
this Book. Return to this question after studying the
following sets of problems.

\end{Itemize}

\NewFolio %% -----File: 158.png---Folio 150-------

\Section{PROBLEMS ON BOOK V}

\begin{Itemize}

\item[1.] What part of the earth's surface lies in the torrid zone?
What part in the temperate zones? What part in the frigid
zones? The parallels $23\tfrac{1}{2}$° north and south of the equator
are the boundaries of the torrid zone, and the parallels $66\tfrac{1}{2}$°
north and south are the boundaries of the frigid zones.

\item[2.] Find to four places of decimals the area of a sphere
circumscribed about a cube whose edge is~$6$. No square root
is to be approximated in the process, and the value of~$\pi$ is
taken as~$3.1416$.

\item[3.] Can the volume of the sphere in the preceding exercise
be approximated without finding a square root? Find the
volume.

\item[4.] Find the area of a sphere circumscribed about a rectangular
parallelopiped whose sides are $a$, $b$, and~$c$.

\item[5.] Find the volume of the sphere in the preceding example.

\item[6.] A fixed sphere with center~$O$ has its center on another
sphere with center~$O'$. Show that the area of the part of~$O'$
which lies within~$O$ is equal to the area
of a great circle of the sphere~$O$, provided
the radius of the sphere~$O$ is not
greater than the diameter of~$O'$.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\begin{Suggestion}
Let the figure represent a cross
section through the centers of the two spheres.
Connect $O$ with $A$ and $B$. Then $\overline{OA}^2 = OB × OD$. But $OD$ is the
altitude of the zone of~$O'$ which lies within~$O$, and $OB$ is the diameter
of the sphere~$O'$. Hence, the area of the zone is $\pi OB × OD = \pi \overline{OA}^2$.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfil\raisebox{0.25in}{\smash{\Input{158a}{1.75in}}}\hfil

\item[7.] Given a solid sphere of radius $12$~inches. A cylindrical
hole is bored through it so that the axis of the cylinder passes
through the center of the sphere. What area of the sphere
is removed if the diameter of the hole is $4$~inches?

\item[8.] Find the volume removed from the sphere by the process
described in the preceding exercise.

\NewFolio %% -----File: 159.png---Folio 151-------

\item[9.] A cylindrical post $6$~in.\ in diameter is surmounted by a
part of a sphere $10$~in.\ in diameter, as shown in the figure.
Find the surface and the volume of the part of
the sphere used.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[10.] A cylindrical post $5$~ft.\ long and $4$~in.\ in
diameter is surmounted by a part of a sphere $9$~in.\
in diameter as shown in the figure. Find
the volume of the whole post including the
spherical part.

\item[11.] Find the volume of a spherical shell
one inch thick if its outer diameter is $8$~inches.

\item[12.] Compare the volumes and areas of a sphere and the circumscribed
cylinder.
\end{minipage}%
% [** Illustration]
\hfill\Input{159a}{1.25in}

\item[13.] In a sphere of radius~$r$ a cylinder is inscribed whose altitude
is equal to its diameter. Compare its volume and area
with those of the sphere.

\item[14.] Find the diagonal of a cube in terms of its side, and also
a side in terms of half the diagonal.

\item[15.] Express the volume of a cube inscribed in a sphere in
terms of the radius of the sphere.

\item[16.] Three spheres each of radius~$r$ are placed on a plane so
that each is tangent to the other two. A fourth sphere of
radius~$r$ is placed on top of them. Find the distance from the
plane to the top of the upper sphere.

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\item[17.] Find the vertical distance from
the floor to the top of a triangular
pile of spherical cannon balls, each
of radius $5$~inches, if there are $3$~layers
in the pile.

\item[18.] Solve a problem like the preceding
if there are $16$~layers in the
pile, each shot of radius~$r$.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{159b}{1.75in}}

\end{Itemize}

\NewFolio %% -----File: 160.png---Folio 152-------

\Section{FORMULAS DEVELOPED IN BOOKS III, IV, V}

\begin{Itemize}

\item[1.] If $V$ is the volume of a rectangular parallelopiped
whose dimensions are $a$, $b$, $c$, then
\[
V = abc.
\]

\item[2.] If $V$ is the volume, $b$ the area of the base, and $h$ the
altitude of a prism or cylinder, then
\[
V = hb.
\]

\item[3.] If $S$ is the lateral surface, $p$ the perimeter of a right
section of a prism or cylinder, and $e$ the lateral edge or element,
then
\[
S = pe.
\]

\item[4.] If $V$ is the volume, $b$ the area of base, and $h$ the altitude
of a pyramid or cone, then
\[
V = \tfrac{1}{3} bh.
\]

\item[5.] If $S$ is the lateral area, $p$ the perimeter of the base, and
$l$ the slant height of a regular pyramid or cone, then
\[
S = \tfrac{1}{2} pl.
\]

\item[6.] If $V$ is the volume, $b$ the lower base, $b'$ the upper base,
and $h$ the altitude of a frustum of pyramid or cone, then
\[
V = \tfrac{1}{3} h (b + b' + \sqrt{bb'}).
\]

\item[7.] If $S$ is the area of the surface, $V$ the volume, and $r$ the
radius of a sphere, then
\[
S = 4 \pi r^2 \text{ and } V = \tfrac{4}{3} \pi r^3.
\]

\item[8.] If $a$ is the spherical excess in degrees of a spherical
polygon, $S$ the area in square units, and $r$ the radius of a
sphere, then
\[
S = \dfrac{a}{720} \cdot 4 \pi r^2.
\]

\item[9.] If $S$ is the area, $h$ the altitude, and $r$ the radius of a lune,
then
\[
S = 2 \pi rh.
\]

\NewFolio %% -----File: 161.png---Folio 153-------

\item[10.] If $V$ is the volume of spherical cone, $h$ the altitude,
and $r$ the radius of the sphere, then
\[
V = \dfrac{2\pi}{3}r^2h.
\]

\item[11.] If $V$ is the volume of a spherical segment, $r$ the radius
of the sphere, $r_1$ and $r_2$ the radii of the bases, and $h$ the altitude
of the segment, then
\[
V = \dfrac{\pi h}{2}(r_1^2 + r_2^2) + \dfrac{\pi}{6} h^3.
\]
\end{Itemize}


\Section{SUPPLEMENTARY EXERCISES IN COMPUTATION}

% [** PP: List items below column-by-column]
\begin{Itemize}

\item[1.] Find the volume of a rectangular parallelopiped if \\
\begin{minipage}{\textwidth}
\noindent\begin{tabular}{cl}
(1) & $a = 6$ \\
    & $b = 8$ \\
    & $c = 7 \tfrac{2}{3}$
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $a = \tfrac{7}{8} $ \\
    & $b = \tfrac{9}{10}$ \\
    & $c = 3$
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $a = \sqrt{3}$ \\
    & $b = \sqrt{6}$ \\
    & $c = 4$
\end{tabular}
\quad
\begin{tabular}{cl}
(4) & $a = 3\sqrt{5}$ \\
    & $b = \sqrt{15}$ \\
    & $c = 3$
\end{tabular}
\end{minipage}%


\item[2.] Find volume of a prism, or cylinder, if \\
\begin{minipage}{\textwidth}
\begin{tabular}{cl}
(1) & $b = 12$ \\
    & $h = 16$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $b = \sqrt{3}$ \\
    & $h = \sqrt{6}$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $b = 4\sqrt{2}$ \\
    & $h = \sqrt{10}$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(4) & $b = 3 \tfrac{1}{4}$ \\
    & $h = 4\sqrt{3}$ \\
\end{tabular}
\end{minipage}%


\item[3.] Find volume of a pyramid or cone if \\
\begin{minipage}{\textwidth}
\begin{tabular}{cl}
(1) & $b = 46$ \\
    & $h = 12$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $b = 47 \tfrac{1}{2}$ \\
    & $h = 10$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $b = 34\pi$ \\
    & $h = 8$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(4) & $b = 42\pi$ \\
    & $h = 6\sqrt{2}$ \\
\end{tabular}
\end{minipage}%


\item[4.] Find the lateral surface of a regular pyramid or cone if \\
\begin{minipage}{\textwidth}
\begin{tabular}{cl}
(1) & $p = 8$ \\
    & $l = 7$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $p = 3\sqrt{7}$ \\
    & $l = \sqrt{21}$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $p = \sqrt{7}$ \\
    & $l = \sqrt{14}$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(4) & $p = 3\sqrt{2}$ \\
    & $l = 4\sqrt{6}$ \\
\end{tabular}
\end{minipage}%


\item[5.] Find the volume of a frustum of a pyramid or cone if \\
\begin{minipage}{\textwidth}
\begin{tabular}{cl}
(1) & $b = 8$ \\
    & $b' = 6$ \\
    & $h = 4$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $b = 2\sqrt{2}$ \\
    & $b' = \sqrt{3}$ \\
    & $h = 6$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $b = 16\pi$ \\
    & $b' = 9\pi$ \\
    & $h = 3$ \\
\end{tabular}
\quad
\begin{tabular}{cl}
(4) & $b = 3\sqrt{2}$ \\
    & $b' = 6\sqrt{3}$ \\
    & $h = 8$
\end{tabular}
\end{minipage}%


\item[6.] Find the volume or surface of a sphere if \\
\null\hfil$r = 4$,\quad $r = 8$,\quad $r = 3\sqrt{2}$, \quad $r = 7\sqrt{3}$, \quad $r = 12\sqrt{6}$\hfil

\item[7.] Find the area of a spherical polygon if \\
\begin{minipage}{\textwidth}
\begin{tabular}{cl}
(1) & $a = 84$° \\
    & $r = 12$
\end{tabular}
\quad
\begin{tabular}{cl}
(2) & $a = 112\text{°}\, 30'$ \\
    & $r = 8$
\end{tabular}
\quad
\begin{tabular}{cl}
(3) & $a = 49\text{°}\, 25'\, 17''$ \\
    & $r = 14$
\end{tabular}
\end{minipage}%
\end{Itemize}

\NewFolio %% -----File: 162.png---Folio 154-------

\Section{MISCELLANEOUS REVIEW EXERCISES}

\subsubsection*{A.~LOCUS PROBLEMS}

\begin{Itemize}
\index{Loci@\textbf{Loci problems}|(}%

\item[1.] Find the locus of all points in space equally distant from
two parallel lines and also from two parallel planes. Discuss.

\item[2.] Find the locus of all points in space equally distant
from each of two intersecting straight lines and also from two
intersecting planes. Discuss.

\item[3.] What is the locus of all points at a perpendicular distance
of $2$~feet from a given line and lying in a plane parallel
to the line? Discuss.

\item[4.] Find the locus of all points equidistant from two given
points $A$ and $B$, and also equidistant from two planes $M$ and
$N$. Discuss.

\item[5.] What is the locus of all points on the floor of a room
which are equally distant from two diagonally opposite corners
of the room, one in the floor and one in the ceiling?

\item[6.] Find the locus of all points on a sphere where it is met
by line-segments of equal length drawn from a fixed point~$P$
outside the sphere. Discuss.

\item[7.] Find the locus of all points which are at the same fixed
distance from each of two intersecting planes, $M$ and $N$, and
also equally distant from two planes, $P$ and $Q$.

\item[8.] Given a plane $M$ and a point $P$ not in $M$. Find the
locus of a point which divides in a given ratio each segment
connecting $P$ with a point in $M$: (a)~if the segments are
divided internally; (b)~if they are divided externally.

\item[9.] A segment $AB$ of fixed length is free to move so that
its end-points lie in two fixed parallel planes. Find the locus
of a point~$C$ on $AB$ if $AC$ is of fixed length.

\item[10.] Given two fixed points in space, through each of which
passes a system of straight lines. If each line of one system
is perpendicular to a line of the other system, find the locus
of the intersection points.

\index{Loci@\textbf{Loci problems}|)}%
\end{Itemize}

\NewFolio %% -----File: 163.png---Folio 155-------

\subsubsection*{B.~PROBLEMS IN NUMERICAL COMPUTATION}

\begin{Itemize}

\item[1.] A pedestal for a monument is in the shape of a frustum
of a regular hexagonal pyramid, the radius of the upper base
being $4$~ft., that of the lower base $6$~ft., and the altitude of the
frustum $8$~ft. Find its volume, slant height, and lateral surface.

\item[2.] The area of the lower base of a frustum of a pyramid is
$42$~sq.~ft., its altitude $8$~ft., and volume $200$~cu.~ft. Find the
area of the upper base.

\item[3.] The area of the base of a pyramid is $480$~sq.~ft.\ and its
altitude $30$~ft. Find the volume of the frustum remaining
after a pyramid with altitude $10$~ft.\ has been cut off by a plane
parallel to the base.

\item[4.] The area of the base of a pyramid is $250$~sq.~in. If a
plane section of the pyramid parallel to the base and at a distance
of $5$~in.\ from it has an area of $175$~sq.~in., find the altitude
of the pyramid.

\item[5.] The figure below represents a solid whose base is a rectangle
$50$~ft.\ long and $40$~ft.\ wide. Its height is $12$~ft.\ and its
top a rectangle $20$~ft.\ by $10$~ft. Find its volume.

% [** Illustration]
\vspace*{-0.25in}
\begin{center}
\Input{163a}{}
\end{center}
\vspace*{-0.25in}

\item[6.] A frustum of a right circular cone has an altitude one
half that of the cone. If its slant height is $8$~ft.\ and lateral
area $64\pi$~sq.~ft., find the diameters of its bases.

\item[7.] If the area of the base of a cone is $16\pi$~sq.~in.\ and its
altitude $6$~in., find the distance from the vertex to a plane,
parallel to the base, which cuts out a section of area $9\pi$~sq.~in.

\end{Itemize}

\NewFolio %% -----File: 164.png---Folio 156-------

\subsubsection*{C.~PROBLEMS IN ALGEBRAIC COMPUTATION}

\begin{Itemize}

\item[1.] Find the total area and the volume of a regular tetrahedron
each of whose edges is~$e$.

\item[2.] If the numerical values of the volume and of the total
area of a regular tetrahedron are equal, what is the length of
its edge?

\item[3.] Find the length of an edge of a regular tetrahedron if
its volume is numerically equal to the square of the edge.

\item[4.] Cut a pyramid of altitude $h$ by means of a plane parallel
to the base so that the perimeter of the section shall be one
third that of the base.\footnoteTrans{This is the declarative form of
Question~5. ---\emph{Trans.}} % [** PP: Added note]

\item[5.] If the altitude of a pyramid is $h$, how far from the base
must a plane parallel to it be drawn so that the area of its
cross section shall be one third that of the base of the pyramid?

\item[6.] In a regular right pyramid a plane parallel to the base
cuts it so as to make a section whose area is one third that of
the base. Find the ratio between the lateral area of the pyramid
and that of the small pyramid cut off by the plane.

\item[7.] If the diameter of a right circular cylinder is equal to
its altitude, determine the diameter so that the total area of
the cylinder shall be equal numerically to its volume.

\item[8.] Cut a right circular cone of altitude $h$ by a plane parallel
to the base so that the area of the section shall be one third
that of the base. Find the distance from the vertex to the plane.

\item[9.] Show that the lateral area of the small cone cut off in
Example~8 is one third the lateral area of the original cone.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[10.] In a right circular cone, with altitude~$h$,
and $r$ the radius of its base, a cylinder
is inscribed as shown in the figure.
Find the radius $OF$ of the cylinder if the
area of the ring bounded by the circles $OF$
and $OA$ is equal to the lateral area of the
small cone cut off at the top.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{164a}{1.25in}}

\end{Itemize}

\NewFolio %% -----File: 165.png---Folio 157-------

\subsubsection*{D.~PROBLEMS IN CONSTRUCTION}

\begin{Itemize}

\item[1.] Construct a plane tangent to a given sphere and parallel
to a given plane. How many such planes are there?

\begin{Suggestion}
From the center of the sphere pass a line perpendicular
to the given plane and draw tangent planes at the points where this line
meets the surface of the sphere. See §§~\Sref{306}, \Sref{129}.
\end{Suggestion}

\item[2.] Construct a plane tangent to a given sphere and perpendicular
to a given line. How many such planes are there?

\item[3.] How many planes may be tangent to a sphere at a point
on the sphere? How many lines? Show how to construct
them.

\item[4.] Through a given point exterior to a sphere construct a
line tangent to the sphere.

\item[5.] How many lines tangent to a sphere can be constructed
from a point outside the sphere?

\item[6.] Through a given point exterior to a sphere construct a
plane tangent to the sphere. How many planes can be passed
through a given exterior point tangent to the sphere?

\item[7.] How many planes tangent to a sphere can be passed
through two given points $A$ and $B$ outside a sphere? Discuss
fully if the line $AB$ (1)~meets the sphere in two points;
(2)~is tangent to the sphere; (3)~does not meet the sphere.
Show how to make the constructions.

\item[8.] Is it possible to construct a spherical triangle each of
whose angles is a right angle? Show how to construct one.
In a trirectangular spherical triangle what is the length of
each side in terms of degrees?

\item[9.] Construct a spherical triangle such that its polar triangle
is identical with the given triangle.

\item[10.] Construct a spherical triangle whose sides are $70$°, $80$°,
and~$110$°, respectively, and find the sides of each of the eight
triangles formed by its polar construction.

\end{Itemize}

\NewFolio %% -----File: 166.png---Folio 158-------

\subsubsection*{E.~THEOREMS TO BE PROVED}

\begin{Itemize}

\item[1.] Prove that planes perpendicular to the faces of a trihedral
angle and bisecting its face angles meet in a line.

\item[2.] Prove that if in two tetrahedrons three faces of one are
equal respectively to three faces of the other and similarly
placed about a vertex, the tetrahedrons are equal.

\item[3.] Prove that two tetrahedrons are equal if two faces and
the included dihedral angle are equal and similarly placed.

\item[4.] In any regular tetrahedron, an altitude equals three times
the perpendicular from its foot to any face;  or an altitude
equals the sum of the perpendiculars to the faces from any
point within the tetrahedron.

\item[5.] In a cylinder of revolution the diameter of whose base
equals the altitude, the volume equals one third the product of
the total surface by the radius of the base.

\item[6.] If two intersecting planes are each tangent to a cylinder,
show that their line of intersection is parallel to an element
of the cylinder and also parallel to the plane containing the
two elements of contact.

\item[7.] The volume of a frustum of a right circular cone equals
the sum of a cylinder and a cone of the same altitude as the
frustum, and with radii which are respectively the half sum
and the half difference of the radii of the frustum.

\item[8.] Of circles on a sphere whose planes pass through a given
point within the sphere, the smallest is that circle whose plane
is perpendicular to the diameter through the given point.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[9.] A right triangular prism is cut by a
plane not parallel to the base, but such
that its intersection $DE$ is parallel to the
base segment~$AB$. Show that the volume
of the part thus cut off is one third the
product of the sum of the three vertical
edges and the area of the base.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{166a}{1.25in}}

\end{Itemize}

\NewFolio %% -----File: 167.png---Folio 159-------

\begin{BioSketch}{Legendre}{Adrien Marie Legendre}
(1752--1833) was a celebrated French mathematician.
He was one of three commissioners who introduced the
metric system in France, having also been a member of the commission
for determining the length of the meter.

Besides many treatises on advanced subjects, he wrote a book on
elementary geometry which was the most successful of the many
attempts to supersede Euclid as a text-book. It went through many
editions in French and was translated into almost every other civilized
language.
\end{BioSketch}

\NewFolio %% -----File: 168.png---Folio 160-------

% not \addcontentsline; no number
\addtocontents{toc}{\protect\contentsline{chapter}%
  {APPENDIX TO SOLID GEOMETRY}{}{}}

\Appendix{I}{SIMILAR SOLIDS}
\SecToc{\texorpdfstring{\protect\makebox[1em][r]{I.} Similar Solids}{Similar Solids}}

\index{Similarity@\textbf{Similarity}|(}

\Paragraph{378.}{Similar Cylinders.} Two right circular cylinders are
\index{Similar@\textbf{Similar}!cylinders of revolution}%{\Pnum{378}}%
\emph{similar} if they are generated by similar rectangles revolving
about corresponding sides.

\begin{Theorem}{379.}{I.}
If in two similar cylinders $s$ and
$s'$ are the lateral areas, $S$ and $S'$ the total areas, $V$ and
$V'$ the volumes, $h$ and $h'$ the altitudes, and $r$ and $r'$ the
radii, then
\[
\dfrac{s}{s'} = \dfrac{S}{S'} = \dfrac{r^2}{r'^2} = \dfrac{h^2}{h'^2} \text{ and } \dfrac{V}{V'} = \dfrac{r^3}{r'^3} = \dfrac{h^3}{h'^3}.
\]
\end{Theorem}

% [** Illustration]
\vspace*{-0.375in}
\begin{center}
\Input{168a}{0.5\textwidth}
\end{center}
\vspace*{-0.25in}

\textbf{Suggestions for Proof}: To find the ratios $\dfrac{s}{s'}$ and $\dfrac{S}{S'}$, make
use of the following, giving reasons for each in detail.
\begin{center}
\begin{tabular}{ll}
(1)\quad $s = 2\pi rh$, & (2)\quad $S = 2\pi r(r + h)$, \\
(3)\quad $\dfrac{r}{r'} = \dfrac{h}{h'}$, \text{ and } & (4)\quad $\dfrac{r + h}{r' + h'} = \dfrac{r}{r'} = \dfrac{h}{h'}$.
\end{tabular}
\end{center}

To find the ratio $\dfrac{V}{V'}$ make use of $V = \pi r^2h$, and $V = \pi {r'}^2h'$,
and $\dfrac{r}{r'} = \dfrac{h}{h'}$.

\NewFolio %% -----File: 169.png---Folio 161-------

\Section{RATIOS RELATING TO SIMILAR CONES}

\Paragraph{380.}{Similar Cones.} Two right circular cones are \emph{similar} if
\index{Similar@\textbf{Similar}!cones of revolution}%{\Pnum{380}}%
they are generated by two similar right triangles revolving
about corresponding sides.

\begin{Theorem}{381.}{II.}
If in two similar cones $s$ and $s'$
are the lateral areas, $S$ and $S'$ the total areas, $V$ and
$V'$ the volumes, $h$ and $h'$ the altitudes, and $r$ and $r'$
the radii, then
\[
\dfrac{s}{s'} = \dfrac{S}{S'} = \dfrac{r^2}{r'^2} = \dfrac{h^2}{h'^2} \text{ and } \dfrac{V}{V'} = \dfrac{r^3}{r'^3} = \dfrac{h^3}{h'^3}.
\]
\end{Theorem}

% [** Illustrations]
\vspace*{-0.5in}
\begin{center}
\Input{169a}{0.4\textwidth}
\end{center}
\vspace*{-0.125in}

\textbf{Suggestions for Proof}: See suggestions under §~\Sref{379}.

\begin{Exercises}

\item[1.] The lateral area of a cone is $36$~square inches. What is
the lateral area of a similar cone whose altitude is $\tfrac{3}{4}$ that of
the given cone?

\item[2.] The total area of one of two similar cones is three times
that of the other. Compare their altitudes and also their
radii.

\item[3.] The sum of the total areas of two similar cones is $144$
square inches. Find the area of each cone if one is $1\tfrac{3}{4}$ times
as high as the other.

\item[4.] The volume of one of two similar cones is $5$~times that
of the other. Compare their altitudes.

\end{Exercises}

\NewFolio %% -----File: 170.png---Folio 162-------

\Section{CONDITIONS FOR SIMILARITY OF TETRAHEDRONS}

\Paragraph{382.}{Similar Polyhedrons.} Two polyhedrons are \emph{similar} if
\index{Corresponding@\textbf{Corresponding}!parts of similar polyhedrons}%{\Pnum{382}}%
\index{Polyhedrons@\textbf{Polyhedrons}!similar}%{\Pnum{382}}%
\index{Similar@\textbf{Similar}!polyhedrons}%{\Pnum{382}}%
they have the same number of faces similar each to each and
similarly placed, and have their corresponding polyhedral angles
equal.

Any two parts which are similarly placed are called \emph{corresponding
parts}, as corresponding faces, edges, vertices.

\begin{Theorem}{383.}{III.}
Two tetrahedrons are similar
if three faces of one are similar respectively to three
faces of the other, and are similarly placed.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{170ab}{0.6\textwidth}
\end{center}

\Objective{Given the tetrahedrons $P-ABC$ and $P'-A'B'C'$ having
$\triangle APB \sim \triangle A'P'B'$, $\triangle APC \sim \triangle A'P'C'$, and $\triangle BPC \sim \triangle B'P'C'$.}
{To prove $P-ABC \sim P'-A'B'C'$.}

\Proof: (1)~Show that $\triangle ABC \sim \triangle A'B'C'$.

(2)~Show that trihedral $\Angles P$ and $P'$ are equal.

Likewise $\angle A = \angle A'$, $\angle B = \angle B'$, $\angle C = \angle C'$.

Hence, by definition, the
polyhedrons are similar.


\noindent\begin{minipage}[b]{\textwidth-2.5in}
\Indent\begin{Sightwork}

\item[1.] If the two prisms in the figure
are similar, name the pairs of corresponding
parts. Likewise for two
similar pyramids.

\end{Sightwork}
\end{minipage}%
% [** Illustrations]
\hfill\raisebox{-0.125in}{\smash{\Input{170cd}{2.25in}}}


\NewFolio %% -----File: 171.png---Folio 163-------

\Section{TETRAHEDRONS HAVING TWO EQUAL TRIHEDRAL ANGLES}

\begin{Theorem}{384.}{IV.}
The volumes of two tetrahedrons,
having a trihedral angle of the one equal to a
trihedral angle of the other, are proportional to the
products of the edges which meet in the vertices of
these angles.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{171ab}{}
\end{center}
\vspace*{-0.125in}

\Objective{Given the tetrahedrons $P-ABC$ and $P'-A'B'C'$ whose volumes
are $V$ and $V'$ and in which Tri.\ $\angle P = {}$ Tri.\ $\angle P'$.}
{To prove that $\dfrac{V}{V'} = \dfrac{PA \cdot PB \cdot PC}{P'A' \cdot P'B' \cdot P'C'}$.}

\Proof: Place $P'-A'B'C'$ so that Tri.\ $\angle P'$ coincides with
Tri.\ $\angle P$.

Let $CM$ and $C'M'$ be the altitudes of $P-ABC$ and $P-A'B'C'$
from the vertices $C$ and $C'$ upon the plane $PAB$.

Let $AN$ and $A'N'$ be the altitudes of the $\triangles PAB$ and $PA'B'$.

Then $\dfrac{V}{V'} = \dfrac{\tfrac{1}{3} CM   \cdot \text{area } PAB}
                            {\tfrac{1}{3} C'M' \cdot \text{area } PA'B'}
  = \dfrac{CM \cdot PB \cdot AN}{C'M' \cdot PB' \cdot A'N'}$.

Now prove $\dfrac{CM}{C'M'} = \dfrac{PC}{PC'}$ and $\dfrac{AN}{A'N'} = \dfrac{PA}{PA'}$.

Hence, we have $\dfrac{V}{V'} = \dfrac{PC \cdot PB \cdot PA}{P'C' \cdot P'B' \cdot P'A'}$.

Give all the steps and reasons in detail.

\NewFolio %% -----File: 172.png---Folio 164-------

\Section{VOLUMES OF SIMILAR TETRAHEDRONS}

\begin{Theorem}{385.}{V.}
The volumes of two similar
tetrahedrons are in the same ratio as the cubes of their
corresponding edges.
\end{Theorem}

\Objective{Given $P-ABC \sim P'-A'B'C'$, with volumes $V$ and $V'$.}
{To prove that $\dfrac{V}{V'} = \dfrac{\overline{PA}^3}{\overline{P'A'}^3}$.}

\Proof: We have $\dfrac{V}{V'} = \dfrac{PA \cdot PB \cdot PC}{P'A' \cdot P'B' \cdot P'C'}$. \hfill §~\Sref{384}

Now use the properties of similar triangles to complete the
proof. Use the figure of §~\Sref{384}.

\begin{Exercises}

\item[1.] Show that two tetrahedrons are similar if they have a
dihedral angle of one equal to a dihedral angle of the other
and the including faces similar each to each  and similarly
placed.

\item[2.] Show that the total areas of two similar tetrahedrons are
in the same ratio as the squares of any two corresponding
edges.

\item[3.] Show that if each of two polyhedrons is similar to a
third they are similar to each other.

\item[4.] Two similar tetrahedral  mounds have a pair of corresponding
dimensions $3$~ft.\ and $4$~ft. If one mound contains
$40$~cu.~ft.\ of earth, how much does the other contain?

\item[5.] The edges of a tetrahedron are $3$, $4$, $5$, $6$, $7$, and~$8$. Find
the edges of a similar tetrahedron containing $64$ times the
volume.

\item[6.] Find what fraction of the altitude of a tetrahedron must
be cut off by a plane parallel to the base, measuring from the
vertex, in order that the new pyramid thus detached may have
one fifth of the original volume.

\end{Exercises}

\NewFolio %% -----File: 173.png---Folio 165-------

\Section{FIGURES HAVING A CENTER OF SIMILITUDE}

\Paragraph{386.}{Center of Similitude.} Two figures are said to have a
\index{Center@\textbf{Center}!of similitude}%{\Pnum{386}}%
\index{Similitude@\textbf{Similitude}!center of}%{\Pnum{386}}%
\index{Symmetry@\textbf{Symmetry}!with respect to a point}%{\Pnum{386}}%
\emph{center of similitude}~$O$, if for any two points, $A$ and $B$, of the
one figure the lines $AO$ and $BO$ meet the other in two points,
$A'$ and $B'$, called \emph{corresponding points}, such that
\[
OA : OA' = OB : OB'.
\]

See figures under §§~\Sref{387}--\Sref{392}.

\begin{Theorem}{387.}{VI.}
\index{Corresponding@\textbf{Corresponding}!linear dimensions}%{\Pnum{387}}%
Any two figures which have a
center of similitude are similar.
\end{Theorem}

\Proof: \Case{(1)}{Two triangles.}

Given\hfil $\dfrac{OA}{OA'} = \dfrac{OB}{OB'} = \dfrac{OC}{OC'}$.\hfil

% [** Illustration]
\vspace*{-0.25in}
\begin{center}
\Input{173a}{}
\end{center}

Let the student prove that $\triangle ABC \sim \triangle A'B'C'$.

In case the triangles do not lie in the same plane, use §~\Sref{101}
to show that the corresponding $\Angles$ are equal.

\Case{(2)}{Two polygons.}

Given \hfil $\dfrac{OA}{OA'} = \dfrac{OB}{OB'} = \dfrac{OC}{OC'}$, etc.\hfil

% [** Illustration]
\begin{center}
\Input{173b}{}
\end{center}

Give the proof both for polygons in the same plane and
not in the same plane.

\NewFolio %% -----File: 174.png---Folio 166-------

\Case{(3)}{Two tetrahedrons.}

% [** Illustration]
\begin{center}
\Input{174a}{}
\end{center}

With the same hypothesis as before, we must prove
% [** PP: Missing triangle symbol added in first forumla]
$\triangle PAB \sim \triangle P'A'B'$,
$\triangle PBC \sim \triangle P'B'C'$,
$\triangle PCA \sim \triangle P'C'A'$, and
then use §~\Sref{383}.

\Case{(4)}{Any two polyhedrons.}

(a)~Prove corresponding polygonal faces similar.

(b)~Prove corresponding polyhedral angles equal.

The last step requires not only equal \emph{face angles} about the
vertex, as in the case of the tetrahedron, but also equal \emph{dihedral}
angles. Note that two dihedral angles are equal if their faces
are parallel right face to right face and left face to left face.
(Why?)

(5)\quad Consider \emph{any two figures whatsoever} having a center of
similitude.

(a)~Take any three points $A$, $B$, $C$, in one figure and the
three corresponding points, $A'$, $B'$, $C'$, in the other.

Then $AB$ and $A'B'$, $AC$ and $A'C'$, etc., are called \emph{corresponding
linear dimensions}, and the triangles $ABC$ and $A'B'C'$
are \emph{corresponding triangles}.

(b)~It is clear that any two corresponding linear dimensions
have the same ratio as any other two, and that any two corresponding
triangles are similar.

In this sense any two figures having a center of similitude
are thus proved to be \emph{similar}.

\NewFolio %% -----File: 175.png---Folio 167-------

\Section{SIMILAR TRIANGLES PLACED IN SIMILITUDE}

\Paragraph{388.}{Ratio of Similitude.} The \emph{ratio of similitude} of two
\index{Ratio@\textbf{Ratio}!of similitude}%{\Pnum{388}}%
\index{Similitude@\textbf{Similitude}!ratio of}%{\Pnum{388}}%
similar figures is the common ratio of their corresponding
linear dimensions. This ratio is the same as the \emph{distance ratio}
of corresponding points from the center of similitude.

\begin{Theorem}{389.}{VII.}
Two similar triangles may be
so placed as to have a center of similitude.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{175a}{}
\end{center}

\Objective{Given the similar triangles $T$ and $T'$, in which
$\dfrac{A'B'}{AB} = \dfrac{A'C'}{AC} = \dfrac{B'C'}{BC}$.}
{To prove that they may be placed with a center of similitude.}

\Proof: From any point $O$ draw $OA$, $OB$, $OC$.

On these rays take $A_1$, $B_1$, $C_1$ so that
\[
\dfrac{OA_1}{OA} = \dfrac{OB_1}{OB} = \dfrac{OC_1}{OC} = \dfrac{A'B'}{AB}.
\]

Now show the following:

(1)\qquad\qquad $\triangle T_1 \sim \triangle T$, and hence $\triangle T_1 \sim \triangle T'$.

(2)\qquad\qquad $\triangle T_1 = \triangle T'$.

For this show that $A_1B_1 = A'B'$ by means of the equations
\[
\dfrac{A_1B_1}{AB} = \dfrac{OA_1}{OA} \text{ and } \dfrac{A'B'}{AB} = \dfrac{OA_1}{OA}.
\]

\NewFolio %% -----File: 176.png---Folio 168-------

Likewise $A_1C_1 = A'C'$ and $B_1C_1 = B'C'$.

(3)~Finally, $\dfrac{OM_1}{OM} = \dfrac{OA_1}{OA}$,
where $M$ and $M_1$ are any two corresponding points.

Hence $O$ is the required center of similitude.

\Section{SIMILAR TETRAHEDRONS PLACED IN SIMILITUDE}

\begin{Theorem}{390.}{VIII.}
Two similar tetrahedrons may
be so placed as to have a center of similitude.
\end{Theorem}

% [** Illustrations]
\begin{center}
\Input{176ab}{}
\end{center}

\Objective{Given the similar tetrahedrons $T$ and $T'$.}
{To prove that they can be placed so as to have a center of
similitude.}

\Proof: With $O$ as a center of similitude, construct $T_1$,
making
\[
\dfrac{OA}{OA_1} = \dfrac{OB}{OB_1} = \text{etc.} = \dfrac{AB}{A'B'}.
\]

Now show as in §~\Sref{389} that $T_1 = T'$, and hence that $T'$ can
be placed in the position $T_1$ so as to have with $T$ the center
of similitude~$O$.

Give all the steps in detail.

\begin{Corollary}{391.}{}
Any two similar polyhedrons may be
placed so as to have a center of similitude.
\end{Corollary}

\begin{Suggestion}
The argument is precisely similar to that in §~\Sref{390}.
\end{Suggestion}

\NewFolio %% -----File: 177.png---Folio 169-------

\Section{CENTER OF SIMILITUDE WITHIN THE FIGURE}

\Paragraph{392.}{} In the proofs for the two preceding theorems the center
of similitude was taken \emph{between} the two figures or \emph{on the same
side} of them. The center may be taken equally well \emph{within}
them, as in the following illustrations:

% [** Illustrations] Fig. 1   Fig. 2
\begin{center}
\Input{177ab}{\textwidth}
\end{center}

In the case of similar convex polyhedrons with the center
of similitude thus placed, the faces are the bases of pyramids
whose vertices are all at the center of
similitude.

If, further, the polygonal faces be
divided into triangles by drawing their
diagonals, these triangles become the
bases of tetrahedrons, all of whose vertices
are at the center of similitude.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-2.25in}
\Indent Moreover, each inner tetrahedron is
similar to its corresponding outer tetrahedron.
(Why?)

The volumes of the two similar polyhedrons are thus composed
of the sums of sets of similar tetrahedrons.
\end{minipage}%
% [** Illustration] Fig. 3
\hfill\smash{\raisebox{-0.75in}{\Input{177c}{2in}}}
\medskip

\begin{Sightwork}

\item[1.] Give the proof of §~\Sref{389}, using \textsc{Fig}.~1 above, and extend the argument
to two similar polygons, using \textsc{Fig}.~2.

\item[2.] Draw two similar tetrahedrons with their center of similitude
within them, and give the proof of §~\Sref{390} with the figure thus made.

\item[3.] Give the proof of §~\Sref{390}, using \textsc{Fig}.~3 above.
\end{Sightwork}

\NewFolio %% -----File: 178.png---Folio 170-------

\Section{RATIO OF VOLUMES OF SIMILAR POLYHEDRONS}

\begin{Theorem}{393.}{IX.}
The volumes of any two similar
polyhedrons are proportional to the cubes of their corresponding
edges.
\end{Theorem}

\Proof: Place the polyhedrons whose volumes are $V$ and $V'$
so as to have their center of similitude within them as in the
third figure of §~\Sref{392}.

Call the volumes of the similar tetrahedrons $T_1$, $T_2$, $T_3$, $\dotsc$,
and $T'_1$, $T'_2$, $T'_3$, $\dotsc$, and let $AB$ and $A'B'$ be two corresponding
edges.

Then we have
\[
\dfrac{\overline{AB}^3}{\overline{A'B'}^3} = \dfrac{T_1}{T'_1} = \dfrac{T_2}{T'_2} = \dfrac{T_3}{T'_3} = \dotsb. \tag*{(Why?)}
\]
And \qquad $\dfrac{T_1 + T_2 + T_3 + \dotsb }{T'_1 + T'_2 + T'_3 + \dotsb } = \dfrac{T_1}{T'_1} = \dfrac{\overline{AB}^3}{\overline{A'B'}^3}$.\hfill (Why?) \\[4pt]
But $T_1 + T_2 + T_3 \dotsb = V$ and $T'_1 + T'_2 + T'_3 \dotsb = V'$.

Hence,\qquad\qquad $\dfrac{V}{V'} = \dfrac{\overline{AB}^3}{\overline{A'B'}^3}$.

\begin{Corollary}{394.}{}
The volumes of any two similar solids are
proportional to the cubes of any two corresponding linear
dimensions.
\end{Corollary}

This proposition may be rendered evident by noticing that
any two similar three-dimensional figures may be built up to
any degree of approximation by means of pairs of similar
tetrahedrons similarly placed. The proposition then holds
for any two corresponding figures used in the process of
approximation.

Note that the ratio of similitude of two similar figures may
be obtained from the ratio of any pair of their corresponding
linear dimensions.

\NewFolio %% -----File: 179.png---Folio 171-------

\Section{APPLICATIONS OF SIMILARITY}

\Paragraph{395.}{The Pantograph.} The theorem that any two figures
\index{Pantograph@\textbf{Pantograph}}%{\Pnum{395}}%
which have a center of similitude are similar is the geometric
basis of many mechanical contrivances for enlarging or reducing
both plane and solid figures; that is, for constructing figures
similar to given figures
and having with them a given
ratio of similitude.

\noindent\begin{minipage}[b]{\textwidth-2.5in}
\Indent The essential property of all
such contrivances is that one
point $O$ is kept fixed, while
two points $A$ and $B$ are allowed
to move so that $O$, $A$, and $B$
always remain in a straight
line, and so that the ratio
$OA : OB$ remains the same. See page~216 of Plane Geometry.
\end{minipage}%
% [** Illustration]
\hfil\smash{\Input{179a}{2in}}\hfil

In the first figure on this page $O$ is a fixed point. Segments
$OD$, $CB$, and the sides of the parallelogram $ACED$ are
of fixed length.

Prove that if $B$ is once so taken on the line $EC$ as to be
in the line $OA$, the points $O$, $A$, and $B$ will always remain
collinear, and that $OA : OB$ remains
a fixed ratio.

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-2.5in}
\Indent In the second figure is shown
an ordinary \emph{pantograph} used
for copying and at the same
time for reducing or enlarging
maps, designs, etc. The lengths
of the various segments are
adjustable, as shown, thus obtaining
any desired scale.
\end{minipage}%
% [** Illustration]
\hfil\smash{\Input{179b}{2in}}\hfil

The same contrivance may be used for copying figures in
\emph{space}, such as relief maps, and at the same time reducing or
enlarging them.
\index{Similarity@\textbf{Similarity}|)}

\NewFolio %% -----File: 180.png---Folio 172-------

\Section{RATIOS RELATING TO ANY TWO SIMILAR SOLIDS}

\Paragraph{396.}{Corresponding Cross Sections.} Now consider any two
\index{Corresponding@\textbf{Corresponding}!cross-sections}%{\Pnum{396}}%
\index{Corresponding@\textbf{Corresponding}!linear dimensions}%{\Pnum{396}}%
\index{Similar@\textbf{Similar}!figures}%{\Pnum{396}}%
similar figures whatever so placed as to have a center of similitude
$O$. We have seen that if points $A$, $B$ and $A'$, $B'$ are corresponding
points of the two figures, then the ratio of the
\emph{corresponding linear dimensions} $AB$ and $A'B'$ is equal to the
ratio of similitude $m : n$ of the two figures.

Also if $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are corresponding
points, then $\triangles ABC$ and $A'B'C'$, and the tetrahedrons $ABCD$
and $A'B'C'D'$ are similar, and we have
\[
\dfrac{\text{area } ABC}{\text{area } A'B'C'} = \dfrac{m^2}{n^2} \text{ and }
\dfrac{\text{vol.\ } ABCD}{\text{vol.\ } A'B'C'D'} = \dfrac{m^3}{n^3}.
\]

The points $A$, $B$, $C$ and $A'$, $B'$, $C'$ determine two planes, each
of which intercepts a certain plane figure in the solid figure
to which the points belong. These two plane figures we call
\emph{corresponding cross sections}.

We assume without full argument:

\begin{Theorem}{397.}{X.}
\Inum{(1)}~The ratio of the areas of
any pair of corresponding cross sections or any pair
of corresponding surfaces of similar figures is equal
to the square of their ratio of similitude, {\upshape and}

\Inum{(2)}~The ratio of the volumes of any two similar figures
is equal to the cube of their ratio of similitude.
\end{Theorem}

The fact that the ratio of the areas of corresponding surfaces
of similar solids is equal to the \emph{square} of their ratio of similitude,
while the ratio of their volumes equals the \emph{cube} of this
ratio is one of the most important and far-reaching conclusions
of geometry.

Thus the ratio of the weights of two similar shells used in
gunnery, or the ratio of the weights of two men of similar
build, may be found when their ratio of similitude is known.

\NewFolio %% -----File: 181.png---Folio 173-------

\Section{PROBLEMS AND APPLICATIONS}

\Page{170}%
\begin{Itemize}

\item[1.] If it is known that a steel wire of radius $r$ will carry a
certain weight~$w$, how great a weight will a wire of the same
material carry if its radius is~$2r$?

\begin{Suggestion}
The tensile strengths of wires are in the same ratio as
their cross-section areas.
\end{Suggestion}

\item[2.] Find the ratio of the diameters of two wires of the same
material if one carries twice the load of the other; three times
the load.

\item[3.] In a laboratory experiment a heavy iron ball is suspended
by a steel wire. In suspending another ball of twice
the diameter a wire of twice the radius of the first one is used.
Is this perfectly safe if it is known that the first wire will just
safely carry the ball suspended from it? Discuss fully.

\item[4.] In two schoolrooms of the same shape (similar figures)
but of different size, the same \emph{proportion} of the floor space is
occupied by desks. Which contains the larger amount of air
for each pupil?

\item[5.] If the shells used in guns are similar in shape, find the
ratio of the total surface areas of an eight-inch and a twelve-inch
shell.

\item[6.] Find the ratio of the weights of the shells in the preceding
problem, weights being in the same ratio as the volumes.

\item[7.] If a man $5$~ft.\ $9$~in.\ tall weighs $165$~lb., what should be
the weight of a man $6$~ft.\ $1$~in.\ tall, if they are similar in shape?

\item[8.] What is the diameter of a gun which fires a shell weighing
twice as much as a shell fired from an eight-inch gun, supposing
the shells to be similar bodies?

\item[9.] Supposing two trees to be similar in shape, what is the
diameter of a tree whose volume is three times that of one
whose diameter is $2$~feet? What is the diameter if the volume
is five times that of the given tree? What if it is $n$~times
that of the given tree?

\NewFolio %% -----File: 182.png---Folio 174-------

\item[10.] Assuming that the weights of schoolboys vary as the
cubes of their heights, construct a graph representing the
relation between their heights and weights, if a boy $5$~feet $9$~inches
tall weighs $130$~pounds.

\begin{Suggestion}
If $w$ represents the number of pounds in weight and $h$
the number of feet in height, $w = kh^3$. From $w = 130$, when $h = 5 \tfrac{3}{4}$, we
have $k = .684$. For the purpose of the graph, $k = .7$ is accurate enough.
\end{Suggestion}

\item[11.] From the graph constructed in the preceding example
find the weight of a boy $5$~feet tall; one $5$~feet $4$~inches; one
$5$~feet $6$~inches. Compare with the weights of boys in your
class.

\item[12.] If a man $6$~feet tall weighs $185$~pounds, construct a graph
representing the weights of men of similar build and of various
heights.

\item[13.] If steamships are of the same shape, their tonnages vary
as the cubes of their lengths. The \emph{Mauretania} is $790$~feet long,
with a net tonnage of~$32,500$. Construct a graph representing
the tonnage of steamships of the same shape, and of various
lengths.

Other ships which at one time or another have held ocean
records are: the (former) \emph{Deutschland}, length $686$~ft.\ and tonnage
$16,500$; the (former) \emph{Kaiser Wilhelm der Grosse}, length
$648$~ft.\ and tonnage $14,300$; the \emph{Lucania}, length $625$~ft.\ and
tonnage $13,000$ (nearly); and the \emph{Etruria}, length $520$~ft.\ and
tonnage $8000$. By means of this graph decide whether or not
these boats have greater or less tonnage than the \emph{Mauretania}
as compared with their lengths.

\item[14.] Raindrops as they start to fall are extremely small. In
the course of their descent a great many are united to form
larger and larger drops. If $1000$ such drops unite into one,
what is the ratio of the surface of the large drop to the sum
of the surfaces of the small drops?

\NewFolio %% -----File: 183.png---Folio 175-------

\item[15.] The strength of a muscle varies as its cross-section area,
which in turn varies as the square of the height or length of
an animal, while the weight of the animal varies as the cube of
its height or length. Use these facts to explain the greater
agility of small animals. For example, compare the rabbit
and the elephant.

\item[16.] Assuming the velocities the same, the amounts of water
flowing through pipes vary directly as their cross-section areas.
How many pipes, each $4$~in.\ in diameter, will carry as much
water as one pipe $12$~in.\ in diameter?

\item[17.] What must be the diameter of a cylindrical conduit
which will carry enough water to supply ten circular intakes
each $8$~feet in diameter?

\item[18.] A water reservoir, including its feed pipes, is replaced
by another, each of whose linear dimensions is twice the corresponding
dimensions of the first. If the velocity of the
water in the feed pipes of the new system is the same as that
in the old, will it take more or less time to fill the new reservoir
than it did the old? What is the ratio of the new time
to the old?

\item[19.] If two engine plants are exactly similar in shape, but
each linear dimension in one is three times the corresponding
dimension of the other, and if the steam in the feed pipes flows
with the same velocity in both, compare the speeds of the
engines.

\item[20.] If two men, one $5$~ft.\ $6$~in.\ and the other $6$~ft.\ $2$~in.\ in
height, are similar in structure in every respect, how much
faster must the blood flow in the larger person in order that
the body tissues of both shall be supplied equally well?

\begin{Suggestion}
Note that the amount of tissue to be supplied varies as
the cube of the height, while the cross-section area of the arteries varies
as the square of the height.
\end{Suggestion}
\end{Itemize}

\Page{172}%

\NewFolio %% -----File: 184.png---Folio 176-------

\Appendix{II}{PROJECTION OF LINE-SEGMENTS}
\SecToc{\texorpdfstring{\protect\makebox[1em][r]{II.} Applications of Projection}{Applications of Projection}}

\Paragraph{398.}{Length of Projection.} The projection of a line-segment
\index{Plane@\textbf{Plane}!projections upon}%{\Pnum{398}}%
\index{Projection@\textbf{Projection}!of a line-segment}%{\Pnum{398}}%
on a plane was shown in §~\Sref{121} to be another line-segment.
The \emph{length of this projection} will now be computed in terms of
the given line-segment.

\Paragraph{399.}{Cosine of Projection Angle.} The acute angle between a
\index{Angle@\textbf{Angle}!of projection}%{\Pnum{399}}%
\index{Projection angle@\textbf{Projection angle}}%{\Pnum{399}}%
line-segment and a given line on which it is projected is called
the \emph{projection angle}.

If $l$ is the length of a line-segment and $p$ the length of its
projection, then the ratio $p:l$ is called the \emph{cosine of the projection
angle}.

\textit{E.g.}, in \textsc{Fig}.~1, $\dfrac{p}{l} = \text{cosine}\, \angle BAE$.

% [** Illustrations] Fig. 1  Fig. 2
\begin{center}
\Input{184ab}{\textwidth}
\end{center}

\Paragraph{400.}{Sine, Cosine, Tangent.} In any right triangle $ABC$
\index{Cosine@\textbf{Cosine of an angle}}%{\Pnum{400}}%
\index{Sine@\textbf{Sine of an angle}}%{\Pnum{400}}%[** PP: Embolden entry]
\index{Tangent@\textbf{Tangent}!of an acute angle}%{\Pnum{400}}%
(\textsc{Fig}.~2), either acute angle, as $\angle A$, is the projection angle
between the hypotenuse and the side adjacent to the angle.

Hence the \emph{cosine} of an acute angle of a right triangle is the
\emph{ratio of the adjacent side to the hypotenuse}.

Likewise we define the \emph{sine} of an acute angle of a right triangle
as the \emph{ratio of the opposite side to the hypotenuse}, and the
\emph{tangent} of an acute angle of a right triangle as the \emph{ratio of the
opposite side to the adjacent side}.

Using the common abbreviations, $\sin$, $\cos$, and $\tan$, we have
in \textsc{Fig}.~2,
\[
\sin A = \dfrac{a}{c},\quad \cos A = \dfrac{b}{c},\quad \tan A = \dfrac{a}{b}. % [** PP: Added period]
\]

\NewFolio %% -----File: 185.png---Folio 177-------

\Section{EXERCISES ON SINES, COSINES, TANGENTS}

The sine, cosine, and tangent are of great importance in
many computations. By careful measurement (and in other
ways) their values may be computed for any acute angle, and
a table formed, like that on page~\Pgref{175}.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\textit{E.g.}, if $\angle A = 35$° (measured with a protractor), and if we measure
$AC$, $AB$, and $BC$, and then compute the ratios
$\dfrac{a}{c}$, $\dfrac{b}{c}$, and $\dfrac{a}{b}$, we shall find the values of $\sin 35$°,
$\cos 35$°, $\tan 35$°.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{185a}{1.5in}}

With an ordinary ruler it will not usually be
possible to make these measurements with sufficient
accuracy to obtain more than one decimal
place.

\begin{Exercises}

\item[1.] Using a protractor, construct angles of $10$°, $30$°, $50$°, $70$°,
and by measurement determine the sine, cosine, and tangent
of each.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\item[2.] Prove that the cosine of any
given angle is the same, no matter
what point is taken in \emph{either} side
from which to let fall the perpendicular
to the other side. Prove the same for the tangent.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{185b}{1.75in}}

\item[3.] Show that if the hypotenuse be taken \emph{one decimeter} in
length, then the length of the side adjacent, measured in
decimeters, is the \emph{cosine} of the angle, and the length of the
side opposite is the \emph{sine} of the angle.

\item[4.] Show that if the side adjacent be taken one decimeter in
length, the length of the side opposite, measured in decimeters,
is the \emph{tangent} of the angle.

\item[5.] Without any direct measurement, show how to compute
the three ratios for each of the angles $30$°, $45$°, $60$°.

\begin{Suggestion}
Make use of the fact that if one acute angle in a right
triangle is $30$°, the side opposite it is one half the hypotenuse.
\end{Suggestion}
\end{Exercises}

\vfil

\NewFolio %% -----File: 186.png---Folio 178-------

\Section{TABLE OF SINES, COSINES, AND TANGENTS}

\index{Table@\textbf{Table of Sines, Cosines, and Tangents}}%{\Pnum{}}%
\Page{175}%
\begin{center}
\footnotesize
\begin{tabular}{%
  |>{\ \bfseries}r@{\quad}|@{\quad}r@{\quad}|@{\quad}r@{\quad}|@{\quad}r@{\quad}|%
  |>{\ \bfseries}r@{\quad}|@{\quad}r@{\quad}|@{\quad}r@{\quad}|@{\quad}r@{\quad}|}
\hline
Angle &  Sin &  Cos  &    Tan &    Angle &  Sin &   Cos &   Tan \\
\hline
 0°  &  0    & 1.000  &  0    &   46°  &  .719  & .695  & 1.04 \\
 1°  & .017  & 1.000  & .017  &   47°  &  .731  & .682  & 1.07 \\
 2°  & .035  &  .999  & .035  &   48°  &  .743  & .669  & 1.11 \\
 3°  & .052  &  .999  & .052  &   49°  &  .755  & .656  & 1.15 \\
 4°  & .070  &  .998  & .070  &   50°  &  .766  & .643  & 1.19 \\
 5°  & .087  &  .996  & .087  &   51°  &  .777  & .629  & 1.23 \\
 6°  & .105  &  .995  & .105  &   52°  &  .788  & .616  & 1.28 \\
 7°  & .122  &  .993  & .123  &   53°  &  .799  & .602  & 1.33 \\
 8°  & .139  &  .990  & .141  &   54°  &  .809  & .588  & 1.38 \\
 9°  & .156  &  .988  & .158  &   55°  &  .819  & .574  & 1.43 \\
10°  & .174  &  .985  & .176  &   56°  &  .829  & .559  & 1.48 \\
11°  & .191  &  .982  & .194  &   57°  &  .839  & .545  & 1.54 \\
12°  & .208  &  .978  & .213  &   58°  &  .848  & .530  & 1.60 \\
13°  & .225  &  .974  & .231  &   59°  &  .857  & .515  & 1.66 \\
14°  & .242  &  .970  & .249  &   60°  &  .866  & .500  & 1.73 \\
15°  & .259  &  .966  & .268  &   61°  &  .875  & .485  & 1.80 \\
16°  & .276  &  .961  & .287  &   62°  &  .883  & .469  & 1.88 \\
17°  & .292  &  .956  & .306  &   63°  &  .891  & .454  & 1.96 \\
18°  & .309  &  .951  & .325  &   64°  &  .899  & .438  & 2.05 \\
19°  & .326  &  .946  & .344  &   65°  &  .906  & .423  & 2.14 \\
20°  & .342  &  .940  & .364  &   66°  &  .914  & .407  & 2.25 \\
21°  & .358  &  .934  & .384  &   67°  &  .921  & .391  & 2.36 \\
22°  & .375  &  .927  & .404  &   68°  &  .927  & .375  & 2.48 \\
23°  & .391  &  .921  & .424  &   69°  &  .934  & .358  & 2.61 \\
24°  & .407  &  .914  & .445  &   70°  &  .940  & .342  & 2.75 \\
25°  & .423  &  .906  & .466  &   71°  &  .946  & .326  & 2.90 \\
26°  & .438  &  .899  & .488  &   72°  &  .951  & .309  & 3.08 \\
27°  & .454  &  .891  & .510  &   73°  &  .956  & .292  & 3.27 \\
28°  & .469  &  .883  & .532  &   74°  &  .961  & .276  & 3.49 \\
29°  & .485  &  .875  & .554  &   75°  &  .966  & .259  & 3.73 \\
30°  & .500  &  .866  & .577  &   76°  &  .970  & .242  & 4.01 \\
31°  & .515  &  .857  & .601  &   77°  &  .974  & .225  & 4.33 \\
32°  & .530  &  .848  & .625  &   78°  &  .978  & .208  & 4.70 \\
33°  & .545  &  .839  & .649  &   79°  &  .982  & .191  & 5.14 \\
34°  & .559  &  .829  & .675  &   80°  &  .985  & .174  & 5.67 \\
35°  & .574  &  .819  & .700  &   81°  &  .988  & .156  & 6.31 \\
36°  & .588  &  .809  & .727  &   82°  &  .990  & .139  & 7.12 \\
37°  & .602  &  .799  & .754  &   83°  &  .993  & .122  & 8.14 \\
38°  & .616  &  .788  & .781  &   84°  &  .995  & .105  & 9.51 \\
39°  & .629  &  .777  & .810  &   85°  &  .996  & .087  &11.43 \\
40°  & .643  &  .766  & .839  &   86°  &  .998  & .070  &14.30 \\
41°  & .656  &  .755  & .869  &   87°  &  .999  & .052  &19.08 \\
42°  & .669  &  .743  & .900  &   88°  &  .999  & .035  &28.64 \\
43°  & .682  &  .731  & .933  &   89°  & 1.000  & .017  &57.29 \\
44°  & .695  &  .719  & .966  &   90°  & 1.000  &  0    &      \\
45°  & .707  &  .707  & 1.00  &        &        &       &      \\
\hline
\end{tabular}
\end{center}

\NewFolio %% -----File: 187.png---Folio 179-------

\Section{LENGTH OF PROJECTION OF A LINE-SEGMENT}

\begin{Theorem}{401.}{I.}
The length of the projection of a
line-segment upon a given line is equal to the length of
the line-segment multiplied by the cosine of the projection
angle.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-2.125in}
\Indent\Objective{Given the projection $p$ of the
line-segment $l$ on the line $CD$,
with the projection angle $A$.}
{To prove that $p = l \cos A$.}

\Proof: By definition we have
$\dfrac{p}{l} = \cos A$. Hence, $p = l \cos A$.
\end{minipage}%
% [** Illustration]
\kern -0.375in \smash{\raisebox{-0.25in}{\Input{187a}{2.5in}}}


\begin{Exercises}

\Page{177a}%

\item[1.] Find the cosines of the angles $35\text{°}\, 30'$, $54\text{°}\, 15'$, $15\text{°}\, 45'$.

\begin{Suggestion}
The cosine of $35\text{°}\, 30'$ lies between $\cos 35$° and $\cos 36$°.
We assume that it lies halfway between these numbers. This assumption,
while not quite correct, is very nearly so for small differences of
angles, as in this case, where the total difference is only one degree.
From the table $\cos 35\text{°} = .819$, $\cos 36\text{°} = .809$. The number midway
between these is~$.814$, which we take as the cosine of $35\text{°}\, 30'$.

This process is called \textbf{interpolation}. A similar process is used for
sines and tangents.
\end{Suggestion}

\item[2.] Find the tangents of the angles $25\text{°}\, 20'$, $47\text{°}\, 45'$, $63\text{°}\, 40'$.

\item[3.] Find the angle whose tangent is $1.74$.

{\small\textit{Solution}.\quad From the table we have $\tan 60\text{°} = 1.73$ and $\tan 61\text{°} = 1.80$.
Hence, the required angle must lie between $60$° and~$61$°. Moreover, the
number $1.74$ is one seventh the way from $1.73$ to~$1.80$. Hence, we
assume the angle to lie one seventh the way from $60$° to~$61$°, which gives
$60\text{°} + \tfrac{1}{7} \times 1\text{°} = 60\text{°} + 9'$ nearly. The required angle is $60\text{°}\, 9'$.}

\item[4.] Find the angles whose sines are $.276$; $.674$; $.437$.

\item[5.] Find the angles whose cosines are $.940$; $.094$; $.435$.

\item[6.] Find the angles whose tangents are $.781$; $1.41$; $3.64$.

\NewFolio %% -----File: 188.png---Folio 180-------

\Page{177b}%

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[7.] At what angle with the horizontal
must the base of a right circular cylinder
be tilted to make it just topple over if its
diameter is $6$~ft.\ and its altitude $8$~ft.?

\begin{Suggestion}
The center of gravity is at the
middle point~$C$ of the axis of the cylinder. The
base must be tilted so that the line $AC$ becomes
vertical. The required angle is $\angle ACB$.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.25in}{\Input{188a}{1.25in}}

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[8.] The Leaning Tower of Pisa is $179$~feet
high and $31$~feet in diameter. It now leans
so that a plumb line from the top on the
lower side reaches the ground $14$~feet from
the base. At what angle is its side now
inclined from the vertical? At what angle
would its side have to incline from the
vertical before it would topple over?
\end{minipage}%
% [** Illustration] [** PP: \smash last]
\hfill\smash{\raisebox{-0.25in}{\Input{188b}{1.25in}}}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\item[9.] A four-inch hole is cut in a board, and a ball $8$~in.\ in
diameter is made to rest on it. At
what angle must the board be held
so that the ball will just roll out
of the hole?

\begin{Suggestion}
The board must be held
so that the line $OA$ becomes vertical.
\end{Suggestion}
\end{minipage}%
% [** Illustration]
\hfill\Input{188c}{1.75in}

\BeforeSkip
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[10.] Using a ball $8$~inches in diameter, what must be the
radius of the hole in the board of the preceding problem so
that the ball shall just roll out when
the board is inclined at an angle of
$45$° to the horizontal?

\Indent\item[11.] If the figure $ABCD-H$ is a
cube, find each of the following
angles: $\angle ECA$, $\angle AEC$.

Check by using the fact that the sum
of the angles of a triangle is $180$°.
\end{minipage}%
% [** Illustration]
\hfill\Input{188d}{1.5in}

\end{Exercises}

\NewFolio %% -----File: 189.png---Folio 181-------

\Section{ALTITUDE OF OBLIQUE PRISM OR CYLINDER}

\begin{Theorem}{402.}{II.}
The altitude of an oblique prism
or cylinder is equal to an element multiplied by the
cosine of the angle between the plane of the base and
that of a right section.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{189a}{0.4\textwidth}
\end{center}

\Objective{Given an oblique prism or cylinder with base $b$ and right section
$c$, and let $BE$ be a perpendicular between the planes of the bases.}{}

Consider the plane determined by $BE$ and the element $AB$.

This plane is $\perp$ to the plane of $b$ and also to the plane of~$c$.
(Why?)

Hence, it is $\perp$ to the line of intersection of the planes of
$b$ and~$c$. (Why?)

Let this plane cut the planes of $b$ and~$c$ in $GD$ and~$FD$
respectively.

Then $\angle D$ is the measure of the dihedral angle between the
planes of $b$ and~$c$. (Why?)
\begin{flalign*}
&\indent \text{\itshape To prove that} & BE &= AB \cdot \cos D. && \\
&\indent \text{\Proof: We have}        & BE &= AB \cdot \cos \angle ABE. && \tag*{Why?} \\
&\indent \text{But}              & \angle D &= \angle ABE. &&\tag*{Why?} \\
&\indent \text{Hence,}                 & BE &= AB \cdot \cos D. &&
\end{flalign*}

\begin{Corollary}{403.}{}
The dihedral angle between the planes of
the base and a right section of an oblique cylinder or prism is
equal to the angle between an element and the altitude.
\end{Corollary}

\NewFolio %% -----File: 190.png---Folio 182-------

\begin{Exercises}

\item[1.] Given a line-segment $10$~inches long. Find the length
of its projection on a plane if the projection angle is~$20$°. If
the angle is $30$°, $45$°, $60$°, $90$°, $0$°.

\item[2.] A kite string forms an angle of $40$° with the ground.
The distance from the end of the string to a point directly
beneath the kite is $200$~ft. Find the length of the string and
the perpendicular height of the kite.

\item[3.] The altitude of an oblique prism is $15$~inches. Find
the length of an element if it makes an angle of $45$° with the
perpendicular between the bases.

\item[4.] A right section of a cylinder makes an angle of $20$° with
the plane of the lower base. Find the ratio between the
altitude and an element.

\noindent\begin{minipage}[b]{\textwidth-2in}
\Indent\item[5.] Prove that by joining the
middle points of six edges of a
cube, as shown in the figure, a
regular hexagon is formed.

\item[6.] Prove that in the preceding
example the plane of the
regular hexagon, $KLMNOP$, is
perpendicular to the diagonal
$DF$ of the cube.
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.125in}{\Input{190a}{2in}}}

\item[7.] How large a cube will be
required from which to cut a stopper for a hexagonal spout,
each of whose sides is $4$~inches?

\item[8.] In the figure find the angle $KQH$.

\begin{Suggestion}
Let $a$ be a side of the cube. Compute $KH$, $KQ$, and
$HQ$ in terms~$a$. Note that $\angle QKH = \text{rt.\ } \angle$.
\end{Suggestion}

\item[9.] Find the area of the projection of the hexagon $KLMNOP$
on the face $BCGF$. Note that this projection equals the whole
square less $\triangle NCO + \triangle KEL$. See §~\Sref{404}. Find the area of
the hexagon in terms of the side~$a$ of the cube.

\end{Exercises}

\NewFolio %% -----File: 191.png---Folio 183-------

\Section{AREA OF THE PROJECTION OF A PLANE-SEGMENT}

\Paragraph{404.}{Projection of a Plane-Segment.} If from each point in
% Next entry gets indexed differently than in the scanned images
\index{Plane@\textbf{Plane}!-segment}%{\Pnum{404}}%
\index{Projection@\textbf{Projection}!of a plane-segment}%{\Pnum{404}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{404}}%
the boundary of a plane-segment a perpendicular is drawn to
a given plane, the locus of the feet of these perpendiculars will
bound a portion of the plane, which is called the \emph{projection of
the plane-segment} on the given plane.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\begin{Remark}
\text{E.g.}, the plane-segment $A'B'C'$ in the plane
$N$ is the projection of the plane-segment $ABC$
from the plane $M$ upon $N$.
\end{Remark}

\Indent The \emph{angle of projection} is the angle
between the planes $M$ and $N$.
\end{minipage}%
% [** Illustration]
\hfill\smash{\Input{191a}{1.5in}}

\begin{Theorem}{405.}{III.}
The area of the projection of a
plane-segment on a plane is equal to the area of the
plane-segment multiplied by the cosine of the projection
angle.
\end{Theorem}

% [** Illustration]
\begin{center}
\Input{191a}{0.4\textwidth}
\end{center}

\Proof: Let the boundary of the given plane-segment $b$ be
any convex polygon or closed curve.

Using a line perpendicular to the given plane as a \emph{generator},
develop a prismatic or cylindrical surface of which $b$
is a section. The given plane will cut this surface in a
right section whose area we denote by~$c$.

Now cut the surface by a plane parallel to~$b$, forming the
upper base $b'$ of a prism or cylinder whose altitude is~$h$, edge~$e$,
and volume~$V$.

\NewFolio %% -----File: 192.png---Folio 184-------

% [** PP: Extra space on right to accommodate figure below]
Then $c$ is the projection of $b$ upon the given plane, and
$\angle 1 = \angle 2$ is the projection angle.
\begin{flalign*}
&\indent \text{We are to show that} & c &= b \cos \angle 1. &&&& \\
&\indent \text{We know that}        & V &= ce = bh. &&&& \tag*{Why?} \\
&\indent \text{But}                 & h &= e \cos \angle 2. &&&& \\
&\indent \text{Hence,}              & ce &= be \cos \angle 2. &&&& \\
&\indent \text{That is,}            & c &= b \cos \angle 2 = b \cos \angle 1. &&&&
\end{flalign*}

\begin{Remark}
\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\textit{Note}. The foregoing theorem may be proved \emph{directly} in case the
plane-segment is a rectangle with one side parallel
to the line of intersection of the two planes.
In the figure let $S$ be the given rectangle and $S'$
its projection, with $AB \parallel$ to the line of intersection
of the planes in which $S$ and~$S'$ lie, and $\angle 1$
the angle between them.
% [** Illustration]
\end{minipage}%
\hfill\smash{\raisebox{-0.5in}{\Input{192a}{1.375in}}}

\begin{flalign*}
&\indent \text{Then}    & S  &= AB \cdot BC && \\
&        \text{and}     & S' &= A'B' \cdot B'C'. && \\
&\indent \text{But}     & AB &= A'B' && \\
&        \text{and}     & BE &= B'C'. && \tag*{Why?} \\
&\indent \text{But}     & BE &= BC \cdot \cos \angle 1 && \tag*{Why?} \\
&        \text{and}     & S' = A'B' &\cdot B'C' = AB \cdot BC \cos \angle 1. && \\
&\indent \text{That is,}& S' &= S \cos \angle 1. &&
\end{flalign*}

In the case of any plane-segment, rectangles may be inscribed in it
in this position and their number increased indefinitely, so that their sum
will approach more and more nearly to the area of the plane-segment,
and in this way it may be shown to any desired degree of approximation,
that the projection of a plane-segment equals the given plane-segment
multiplied by the cosine of the projection angle.
\end{Remark}

\Paragraph{406.}{Ellipse.} An important special case of the theorem of
\index{Ellipse@\textbf{Ellipse}!area of}%{\Pnum{406}}%
§~\Sref{405} is the area of the figure obtained by projecting a circle
upon a plane not parallel to its own plane, nor at right angles
to it. This figure is called an \emph{ellipse}. On page~\Pgref{182} the principle
developed above is used to find the area of the ellipse.

\NewFolio %% -----File: 193.png---Folio 185-------

\Section{THE AREA OF AN ELLIPSE}

\Paragraph{407.}{Projection of a Circle.} In the figure two planes, $M$ and
\index{Projection@\textbf{Projection}!of a circle}%{\Pnum{407}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{407}}%
$M'$, meet in a line $PQ$. The circle $O$ in $M$ has a diameter
$AB \parallel PQ$ and a diameter $CD \perp PQ$.

In projecting the whole figure upon the plane $M'$ the
diameter $AB$ projects into its equal $A'B'$, while $CD$ projects
into $C'D'$ so that $C'D' = CD \cos \angle 1$.

% [** Illustration]
\begin{center}
\Input{193a}{}
\end{center}

\Page{182}%
By theorem §~\Sref{405} the area of the ellipse $A'C'B'D'$ equals the
area of the circle $ACBD$ multiplied by $\cos \angle 1$.

Hence, $\pi r^2 \cdot \cos \angle 1$ is the area of the ellipse.

But $r \cos \angle 1 = O'C'$ and $r = O'B'$. \hfill (§~\Sref{401})

Hence, the area of the ellipse is $\pi \cdot O'C' \times O'B'$.

The segments $A'B'$ and $C'D'$ are called respectively the
major and minor axes of the ellipse, and $O'B'$ and $O'C'$ the
semimajor and the semiminor axes. These latter are usually
denoted by $a$ and~$b$.

\textbf{Hence,\qquad\qquad the area of the ellipse is $\pi ab$}.
\index{Ellipse@\textbf{Ellipse}!area of}%{\Pnum{407}}%

\begin{Remark}
Note that when $a$ and $b$ are equal, the ellipse becomes a circle, and
this formula reduces to $\pi a^2$ as it should.
\end{Remark}

\NewFolio %% -----File: 194.png---Folio 186-------

\Section{PROBLEMS AND APPLICATIONS}

\Page{183}%
\begin{Itemize}

\item[1.] Given a right circular cylinder the radius of whose base
is $6$~inches. Find the area of an oblique cross section inclined
at an angle of $45$° to the plane of the base.

\item[2.] Given an oblique circular cylinder the radius of whose
right section is $10$~inches. Find the area of the base if it is
inclined at an angle of $60$° to the right section.

\item[3.] If an oblique circular cylinder has an altitude~$h$, an element
$e$, radius of right section~$r$, and $\angle A$ the inclination of
the base to the right section, express the volume in two ways
and show that these are equivalent.

\noindent\begin{minipage}[b]{\textwidth-1.5in}
\Indent\item[4.] A six-inch stovepipe has a $45$° elbow
angle, that is, it turns at right angles. (The
angle $CAB$ is called the elbow angle.) Find
the area of the cross section of $AB$. Likewise
if it has a $60$° elbow angle.

\Indent\item[5.] At what angle must the damper in a
circular stovepipe be turned in order to obstruct just half the
right cross sectional area of the pipe?
\end{minipage}%
% [** Illustration]
\hfill\raisebox{0.25in}{\Input{194a}{1.5in}}

\begin{Suggestion}
The damper must be turned so that the projection of
the damper upon a right cross section is equal to half that cross section.
\end{Suggestion}

\item[6.] The comparatively low temperature of the earth's surface
near the pole, even in summer, when the sun does not set for
months, is due partly to the \emph{obliqueness} with which the sun's
rays strike the earth. That is, a given amount of sunlight is
spread over a larger area than in lower latitudes.

\begin{Remark}
Thus, if in the figure $D'C$ is a horizontal line, and $D'D$ the direction
of the sun's rays, then a beam of light
whose right cross section is $ABCD$ is spread
% [****** PP: Width-dependent line break, set for 5in]

\noindent\begin{minipage}[b]{\textwidth-2.25in}
over the rectangle $A'BCD'$.
In other words, a patch of ground
$A'BCD'$ receives only as much sunlight
as a patch the size of $ABCD$
receives when the sun's rays strike it
vertically. $ABCD = A'BCD' \cos \angle 1$.
\end{minipage}%
% [** Illustration]
\hfil\smash{\Input{194b}{1.75in}}\hfil
\end{Remark}

\end{Itemize}

\NewFolio %% -----File: 195.png---Folio 187-------

\begin{Remark}
Hence, each unit of area in $A'BCD'$ receives $\cos \angle 1$ times as much
light as a unit in $ABCD$.

Hence, to compare the heat-producing powers of sunlight in any latitude
with that at the place where the sun's rays fall vertically, we need
to know how the projection angle, $\angle 1$, is related to the difference in latitude
of the two places.
\end{Remark}

\begin{Itemize}

\item[7.] If $\angle 1 = 30$°, compare the amount of heat received by a
unit of area in $ABCD$ and $A'BCD'$.

\item[8.] What must $\angle 1$ be in order that a unit of area in $A'BCD'$
shall receive only $\tfrac{1}{5}$ as much light as a unit in $ABCD$?

\item[9.] The figure represents a cross section of the earth with an
indication of the direction of the rays of light as they strike
it at the summer solstice when they are vertical at $A$, the
tropic of Cancer. $B$ represents the latitude of Chicago, $C$ the
polar circle, and $P$ the north pole. The angles $PDE$, $CGF$,
$BKH$ represent the projection angle, $\angle 1$, for the various
latitudes. Prove:

\noindent\begin{minipage}[b]{\textwidth-3in}
\begin{align*}
\angle PDE &= \angle POK, \\
\angle CGF &= \angle COK, \\
\angle BKH &= \angle BOK.
\end{align*}

\begin{Remark}
That is, $\angle 1$ for each place
is the latitude of that place
minus the latitude of the
place where the sun's rays
are vertical.
\end{Remark}
\end{minipage}%
% [** Illustration]
\hfil\smash{\Input{195a}{2.75in}}\hfil

\item[10.] Find the relative amount of sunlight received by a unit
of area at the tropic of Cancer and at the north pole at the
time of the summer solstice.

\item[11.] Find the ratio between the amount of light received by
a unit of the earth's area at Chicago and at the tropic of
Cancer at the time of the summer solstice.

\item[12.] Find the same ratio for the equator and Chicago at the
winter solstice when the sun is vertical at latitude $23\tfrac{1}{2}$°~south.
\Page{184}%

\end{Itemize}

\NewFolio %% -----File: 196.png---Folio 188-------

\Appendix{III}{VARIABLES. LIMITS}
\SecToc{\texorpdfstring{\protect\makebox[1em][r]{III.} Theory of Limits}{Theory of Limits}}
\index{Theory@\textbf{Theory of Limits}}%{\Pnum{}}%

\Paragraph{408.}{Variables and Functions.} It is often useful to think of
\index{Function@\textbf{Function}}%{\Pnum{408}}%
\index{Variables@\textbf{Variables}}%{\Pnum{408}}%
a geometric figure as \emph{varying} in size and shape.

\begin{Remark}
\textit{E.g.}, if a rectangle has a fixed base, say $10$~inches long, but an altitude
which varies from $3$~inches to $5$~inches, then the area varies from $3 \cdot 10 = 30$
to $5 \cdot 10 = 50$ square inches.

We may even think of the altitude as starting at zero inches and
increasing, in which case the area starts at zero and increases continuously.
\end{Remark}

The altitude which we think of as varying at our pleasure is
called the \emph{independent variable}, while the area, being dependent
upon the altitude, is called the \emph{dependent variable}.

The dependent variable is sometimes called a \emph{function} of the
independent variable, meaning that the two are connected by
a definite relation such that, for any definite value of the independent
variable, the dependent variable also has a definite
value.

\begin{Remark}
Thus, in the formula for the area of a rectangle, $a = bh$, if $b$ is fixed
and $h$ varies, then $a$ is a function of $h$, since for every value of $h$ there is
determined a definite value of~$a$.
\end{Remark}

\Paragraph{409.}{Illustrations of Limits.} If a regular polygon is inscribed
\index{Limit@\textbf{Limit}!of a variable}%{\Pnum{409}}%
in a circle of fixed radius, and if the number of sides of the
polygon be continually increased, for instance by repeatedly
doubling the number, then the apothem, perimeter, and area
are all variables depending upon the number of sides. That is,
each of these is a function of the number of sides.

Now the greater the number of sides the more nearly does
the apothem equal the radius in length. Indeed, it is evident
that the difference between the apothem and the radius will
ultimately become less than any fixed number, however small.
Hence we say that the apothem \emph{approaches the radius as a limit}
\index{Approach@\textbf{Approach}}%{\Pnum{409}}%
as the number of sides increases indefinitely.

\NewFolio %% -----File: 197.png---Folio 189-------

Similarly the perimeters of the polygons may be made as
nearly equal to the circumference as we please by making the
number of sides sufficiently great.

Hence we may define the circumference of a circle as the
\emph{limit of the perimeter} of a regular inscribed polygon as the
number of sides increases indefinitely.

The circumference of a circle may also be defined as the
limit of the perimeter of a circumscribed polygon as the number
of sides is increased indefinitely.

Likewise we may define the area of a circle as the \emph{limit of
the area} of the inscribed or the circumscribed polygon as the
number of sides is increased indefinitely.

The notion of a limit may be used to define the length of a
line-segment which is incommensurable with a given unit segment.
% [** PP: Keeping wicked which]

\begin{Remark}
Thus, the diagonal $d$ of a square whose side is unity is $d = \sqrt{2}$. Hence
$d$ may be defined as the \emph{limit of the variable line-segment} whose successive
lengths are $1$, $1.4$, $1.41$, $1.414$, $\dotsc$.\footnoteTrans%
  {See \Tref{197a}{Transcriber's Notes}.}
\Erratum{err197a} % [** PP: Bad mathematical definition...]
\end{Remark}

In like manner, the length of any line-segment, whether
commensurable or incommensurable with the unit segment,
may be defined in terms of a limit.

\begin{Remark}
Thus, if a variable segment is increased by successively adding to it
one half the length previously added, then the segment will approach a
limit. If the initial length is $1$, and if the successive additions are $\tfrac{1}{2}$, $\tfrac{1}{4}$, $\tfrac{1}{8}$,
$\tfrac{1}{16}$, $\tfrac{1}{32}$, etc., then the successive lengths are $1$, $1 \tfrac{1}{2}$, $1 \tfrac{3}{4}$, $1 \tfrac{7}{8}$, $1 \tfrac{15}{16}$, $1 \tfrac{31}{32}$, etc.
Evidently this segment approaches the limit~$2$.

Hence $2$ may be defined as the limit of the variable segment whose
successive lengths are $1$, $1 \tfrac{1}{2}$, $1 \tfrac{3}{4}$, $1 \tfrac{7}{8}$, etc., as the number of successive additions
is increased indefinitely.
\end{Remark}

The idea of a functional relation between variables and the
idea of a limit, as illustrated above, are two of the most important
concepts in all mathematics.

\NewFolio %% -----File: 198.png---Folio 190-------

\Section{DEFINITION OF A LIMIT}

\Paragraph{410.}{Constant.} A  quantity  which remains fixed in value
throughout a discussion is called a \emph{constant}.
\index{Constant@\textbf{Constant}}%{\Pnum{410}}%

\begin{Remark}
\textit{E.g.}, the base of the rectangle mentioned in §~\Sref{408} is a constant.
\end{Remark}

\Paragraph{411.}{Variable.} A quantity which continuously changes in
\index{Variables@\textbf{Variables}}%{\Pnum{411}}%
value, or which takes on a succession of different values, is
called a \emph{variable}.

\begin{Remark}
\textit{E.g.}, the altitude and the area mentioned in §~\Sref{408} are variables.
\end{Remark}

\Paragraph{412.}{Limit of a Variable.} If a variable may be made to
\index{Limit@\textbf{Limit}!of a variable}%{\Pnum{412}}%
approach a certain constant quantity in such a way that the
difference between the constant and the variable becomes and
remains less than any assignable value, however small, then
the constant is called the \emph{limit of the variable}.

\begin{Remark}
\textit{E.g.}, the fixed circumference of the circle is the limit of the variable
perimeters of the polygons mentioned in §~\Sref{409}.
\end{Remark}

\begin{Sightwork}

\item[1.] Find the limit of a variable line-segment whose initial length is $6$~inches,
and which varies by successive increments each equal to one half
the preceding, the first increment being $2$~inches.

\item[2.] Construct a right triangle whose sides are $1$ and~$2$. By approximating
a square root, find five successive lengths of a segment which approaches
the length of the hypotenuse as a limit.

\item[3.] If one tangent to a circle is fixed and another is made to move so
that their intersection point approaches the circle, what is the limiting
position of the moving tangent?    What is the limit of the measure of the
angle formed by these tangents?

\item[4.] The arc $AB$ of $74$° is the greater of the two arcs intercepted between
two secants meeting at $C$ outside the circle. The points $A$ and $B$
remain fixed while $C$ moves up to the circle. What angle is the limit of
the variable angle formed by the varying secants? What is the limit
of the measure of this angle?

\item[5.] If in the preceding the secants meet within the circle, what is the
limit of their angle and also of the measure of this angle as the intersection
point moves up to the center of the circle?

\end{Sightwork}

\NewFolio %% -----File: 199.png---Folio 191-------

\Section{THE INCOMMENSURABLE CASES}

\Paragraph{413.}{The Incommensurable Cases.} We have seen that there
\index{Incommensurables@\textbf{Incommensurables}}%{\Pnum{413}}%
are segments which are \emph{incommensurable}; that is, which have
no common unit of measure,---for instance, the side and the
diagonal of a square.

For \emph{practical purposes} the lengths of such segments are approximated
to any desired degree of accuracy, and their ratios
are understood to be the ratios of these approximate numerical
measures.

But for \emph{theoretical purposes} it is important to consider these
incommensurable cases further; just as in algebra we not only
\emph{approximate} such roots as $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, etc., but we also deal
with these surds as \emph{exact} numbers.

\begin{Remark}
Instances of this kind occur in such operations as
\[
(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1.
\]
\end{Remark}

While the length of the diagonal of a unit square cannot be
expressed as an integer or as a rational fraction, that is, as the
quotient of two integers, we nevertheless think of such a segment
as having a \emph{definite length}, or what is the same thing, a
\emph{definite ratio} with the unit segment forming the side of the
square.

We now fix our attention on the \emph{incommensurable ratios
themselves}, and the method of determining them, rather than
on the \emph{process of approach} and the practical computation based
on it.

\Paragraph{414.}{Representation by Irrational Numbers.} In general, the
\index{Irrational@\textbf{Irrational}!number}%{\Pnum{414}}%
ratio between any two incommensurable geometric magnitudes
of the same kind may be represented by what is called an
\emph{irrational number}; that is, a number which is \emph{neither an integer
nor a quotient of two integers}.

Examples of irrational numbers are the surds, such as $\sqrt{2}$, % [** PP: Added comma]
$\sqrt[3]{5}$, etc., and the number~$\pi$.

\NewFolio %% -----File: 200.png---Folio 192-------

\Section{NUMBERS DEFINED BY SEQUENCES}

\Paragraph{415.}{Numbers Defined by Sequences.} The following is a
\index{Approach@\textbf{Approach}}%{\Pnum{415}}%
\index{Sequence@\textbf{Sequence}!numbers defined by}%{\Pnum{415}}%
method for determining any number, whether rational or
irrational. For simplicity it is applied first to the integer~$1$.

Throughout this discussion the expressions ``point on a line''
and ``number'' will be used interchangeably.

\begin{Remark}
\begin{wrapfigure}[3]{r}{2.25in}
% [** Illustration]
\Input{200a}{2in}
\end{wrapfigure}
On a straight line mark a certain point $0$ (zero), and one unit to the
right of it mark another point~$1$. Also
lay off points such that their distances
from $0$ are % [** PP: Remove spurious comma after ``are'']
$\tfrac{1}{2}$, $\tfrac{3}{4}$, $\tfrac{7}{8}$, $\tfrac{15}{16}$, $\dotsc$.

If this sequence of points is carried ever so far, it will never reach the
point~$1$. If, however, we select a point to the left of~$1$, no matter how
near to it, we may always go far enough along this sequence to reach
points between this point and~$1$. That is, $1$~is the limit of the terms of
this sequence. See §~\Sref{412}.
\end{Remark}

The point $1$ has two definite relations to this sequence:

(a)~\textit{Every point of the sequence is to the left of~$1$}.

(b)~\textit{For any fixed point to the left of $1$ there are points of the
sequence between it and~$1$}.

We see that~$1$ is the \emph{only} point on the whole line such that
both (a) and~(b) are true of it. For every point to the right
of~$1$ (a)~is true, but (b)~is not. For every point to the left of~$1$
(b)~is true, but (a)~is not.

It follows therefore that, while the points of the sequence
merely \emph{approach} the point~$1$ as a limit, the sequence, \emph{taken as
a whole, serves to distinguish that point from all other points on
the line}.

\begin{Remark}
In the above sequence, the terms continually \emph{increase}. The number~$1$
may be determined equally well by a \emph{decreasing} sequence. For instance,
the sequence $1 \tfrac{1}{2}$, $1 \tfrac{1}{4}$, $1 \tfrac{1}{8}$, $1 \tfrac{1}{16}$, $\dotsc$ has the following properties which distinguish
the point~$1$ from all other points on the line.
\end{Remark}

(a)~\textit{Every point of the sequence is to the right of~$1$}.

(b)~\textit{For any fixed point to the right of~$1$ there are points of the
sequence between it and~$1$}.

\NewFolio %% -----File: 201.png---Folio 193-------

\Section{LIMIT OF AN INFINITE SEQUENCE}

\Paragraph{416.}{Infinite Sequences.} An endless sequence of either kind
just described is called an \emph{infinite sequence}.

Not every infinite sequence serves to single out a definite
point in the manner shown above. Thus the sequence $1$, $2$, $3$,
$4$, $\dotsc$ fails to do so, because its terms grow large beyond all
bound. Such sequences are said to be \emph{unbounded}, while the
sequences $\tfrac{1}{2}$, $\tfrac{3}{4}$, $\tfrac{7}{8}$, $\dotsc$ and $1 \tfrac{1}{2}$, $1 \tfrac{1}{4}$, $1 \tfrac{1}{8}$, $\dotsc$ are \emph{bounded}.
\index{Sequence@\textbf{Sequence}!bounded}%{\Pnum{416}}%
\index{Sequence@\textbf{Sequence}!infinite}%{\Pnum{416}}%
\index{Sequence@\textbf{Sequence}!oscillating}%{\Pnum{416}}%
\index{Sequence@\textbf{Sequence}!unbounded}%{\Pnum{416}}%
\index{Limit@\textbf{Limit}!of a sequence}%{\Pnum{416}}%

Again, the sequence $1$, $2$, $1$, $2$, $1$, $2$, $\dotsc$ fails to single out a
definite point. This sequence is said to be \emph{oscillating}, since its
terms increase, then decrease, then increase, etc., while $\tfrac{1}{2}$, $\tfrac{3}{4}$,
$\tfrac{7}{8}$, $\dotsc$ and $1 \tfrac{1}{2}$, $1 \tfrac{1}{4}$, $1 \tfrac{1}{8}$, $\dotsc$ are \emph{non-oscillating}.

\Paragraph{417.}{Limit of a Sequence.} The number $1$ is said to be the
\index{Bound@\textbf{Bound}!greatest lower}%{\Pnum{417}}%
\index{Bound@\textbf{Bound}!least upper}%{\Pnum{417}}%
\emph{least upper bound} of the sequence $\tfrac{1}{2}$, $\tfrac{3}{4}$, $\tfrac{7}{8}$, $\dotsc$. That is, $1$ is the
\emph{smallest number} beyond which the sequence does not go. $1$ is
also said to be the \emph{limit of the sequence}. See §~\Sref{412}.

Similarly, $1$ is the \emph{greatest lower bound} or the \emph{limit} of the
sequence $1 \tfrac{1}{2}$, $1 \tfrac{1}{4}$, $1 \tfrac{1}{8}$, $\dotsc$; that is, the \emph{greatest number} such that
the sequence contains no number less than it.
\index{Sequence@\textbf{Sequence}!limit of}%{\Pnum{417}}%

In the manner described above any integer may be determined
by an increasing or a decreasing infinite sequence.

\begin{Remark}
\textit{E.g.}, the number $2$ is the limit of the increasing sequence $1$, $1 \tfrac{1}{2}$, $1 \tfrac{3}{4}$, $1 \tfrac{7}{8}$,
$1 \tfrac{15}{16}$, $\dotsc$, or of the decreasing sequence $2 \tfrac{1}{2}$, $2 \tfrac{1}{4}$, $2 \tfrac{1}{8}$, $2 \tfrac{1}{16}$, $\dotsc$.
\index{Sequence@\textbf{Sequence}!decreasing, increasing}%{\Pnum{417}}%
\end{Remark}

Two different sequences, both increasing or both decreasing,
may also define the same number.

\begin{Remark}
\textit{E.g.}, each of the increasing sequences $\tfrac{1}{2}$, $\tfrac{3}{4}$, $\tfrac{7}{8}$, $\tfrac{15}{16}$, $\dotsc$ and $\tfrac{2}{3}$, $\tfrac{4}{5}$, $\tfrac{6}{7}$, $\tfrac{8}{9}$, $\dotsc$
determines the number~$1$.

We notice, however, that no matter what definite number we select in
either of these sequences, there is a number in the other greater than it;
that is, neither sequence contains a number greater than every number in
the other. See §~\Sref{420}.
\end{Remark}

It is apparent that any number may be defined as the limit
of a sequence, and we shall assume the converse, namely, that
any bounded increasing or decreasing sequence has a limit.
\Page{190}%

\NewFolio %% -----File: 202.png---Folio 194-------

\Section{FUNDAMENTAL PRINCIPLES OF LIMITS}

\begin{Axiom}{418.}{}
\index{Axioms@\textbf{Axioms}}%{\Pnum{418}}%
We now assume that: \textit{Every bounded increasing
sequence has a least upper bound, and every bounded decreasing
sequence has a greatest lower bound}.
\end{Axiom}

This axiom may also be stated:

\textit{Every bounded increasing or decreasing sequence has a limit}.

This axiom simply means that every such sequence singles
out a definite number, rational or irrational, in the manner
discussed on page~\Pgref{190}.

\begin{Remark}
Thus, if we attempt to approximate the square root of $2$, we obtain a
sequence, $1$, $1.4$, $1.41$, $1.414$, $1.4142$, $\dotsc$, having for its limit a \emph{definite
number} represented by $\sqrt{2}$, which corresponds to the length of the diagonal
of a square whose side is unity.

Again, if we attempt to approximate the value of~$\pi$, we obtain a
sequence $3$, $3.1$, $3.14$, $3.141$, $3.1415$, $\dotsc$, having for its limit the definite
number represented by~$\pi$, which is the ratio of the circumference to the
diameter of a circle.
\end{Remark}

\begin{Theorem}{419.}{I.}
Two bounded sequences have the
same limit if they are equal term by term.
\end{Theorem}

\Proof: If the sequences are $a_1$, $a_2$, $a_3$, $\dotsc$ and $b_1$, $b_2$, $b_3$, $\dotsc$, and
if $a_1 = b_1$, $a_2 = b_2$, $a_3 = b_3$, $\dotsc$, then they are one and the same
sequence and hence have the same limit by §~\Sref{418}.

\begin{Remark}
\textbf{Historical Note}.\quad The rigorous treatment of limits has been developed
during the last fifty years. The foundation on which it rests is the theory
of irrational numbers, which itself was placed on a firm basis in the early
seventies.

The Greeks dealt with the incommensurable cases in an interesting
manner. Thus, to prove that two incommensurable ratios are equal they
showed that neither can be less than the other. However, they failed to
make explicit definitions of incommensurables and they did not explicitly
state the fundamental axiom of §~\Sref{418} or its equivalent. Their treatment
would be complete and rigorous if supplemented by the proper definitions
and axioms.
\end{Remark}

\NewFolio %% -----File: 203.png---Folio 195-------

\Section{FUNDAMENTAL PRINCIPLES OF LIMITS}

\begin{Theorem}{420.}{II.}
Two increasing bounded sequences
have the same limit if neither sequence contains a number
greater than every number of the other.
\end{Theorem}

Let $a_1$, $a_2$, $a_3$, $\dotsc$ and $b_1$, $b_2$, $b_3$, $\dotsc$ denote two infinite sequences
with limits $a$ and~$b$, such that no term of the first sequence is
greater than every term of the second, and no term of the
second is greater than every term of the first.

\textit{To prove that $a = b$}.

\Proof: Suppose that $a$ is not equal to $b$ but is less than~$b$.
Then there must be numbers of the sequence $b_1$, $b_2$, $b_3$, $\dotsc$ greater
than $a$, since by §~\Sref{415} there are numbers $b_1$, $b_2$, $b_3$, greater than
any fixed number whatever which is less than~$b$. This contradicts
the hypothesis of the theorem, and hence $a$ cannot be
less than~$b$. In like manner we show that $b$ is not less than~$a$.
Hence $a = b$. See §~\Sref{20}.

\begin{Theorem}{421.}{III.}
Two decreasing bounded sequences
have the same limit if neither sequence contains
a number less than every number of the other.
\end{Theorem}

\Proof: The proof is exactly similar to that of Theorem~II.

\begin{Exercises}

\item[1.] Show that the sequences $\tfrac{1}{2}$, $\tfrac{3}{4}$, $\tfrac{7}{8}$, $\dotsc$ and $\tfrac{2}{3}$, $\tfrac{4}{5}$, $\tfrac{6}{7}$, $\dotsc$ have the
same limits.

\item[2.] Show that the sequences $\tfrac{1}{2}$, $\tfrac{1}{4}$, $\tfrac{1}{8}$, $\dotsc$ and $\tfrac{1}{3}$, $\tfrac{1}{5}$, $\tfrac{1}{7}$, $\dotsc$ have the
same limits.

\end{Exercises}

The applications of the theory of limits to geometry consist
chiefly in showing that two numbers are equal because they are
the limits of the same sequence, or of sequences having the
property stated in the theorem of §~\Sref{420} or of §~\Sref{421}.

\NewFolio %% -----File: 204.png---Folio 196-------

\Section{APPLICATION OF LIMITS TO PLANE GEOMETRY}

\begin{Problem}{422.}
\index{Commensurable@\textbf{Commensurable}!segments}%{\Pnum{422}}%
\index{Sequence@\textbf{Sequence}!approximating}%{\Pnum{422}}%
On a given segment $AB$ to lay off a
sequence of points $B_1$, $B_2$, $B_3$, $\dotsc$, of which $B$ is the limit,
such that each of the segments $AB_1$, $AB_2$, $AB_3$, $\dotsc$ is
commensurable with a given segment~$CD$.
\end{Problem}

% [** Illustrations]
\vspace*{-0.25in}
\begin{center}
\Input{204a}{}
\end{center}
\vspace*{-0.125in}

{\small\textit{Solution}:\quad Using $m_1$, an exact divisor of $CD$, as a unit of measure,
lay off on $AB$ a segment $AB_1$, an exact multiple of $m_1$, such that the
remainder $B_1B$ is less than $m_1$. Then $CD$ and $AB_1$ are commensurable.

Using a unit $m_2$, likewise a divisor of $CD$, and less than $B_1B$, lay off
$AB_2$ such that $B_2B$ is less than $m_2$. Then $AB_2 > AB_1$, and $CD$ and $AB_2$
are commensurable.

Continuing in this manner, using as units of measure segments $m_3$,
$m_4$, $\dotsc$ each an exact divisor of $CD$ and each less than $B_2B$, $B_3B$, $\dotsc$
respectively, we obtain a sequence of segments $AB_1$, $AB_2$, $AB_3$, $\dotsc$, each
greater than the preceding and each commensurable with~$CD$.

If the units $m_1$, $m_2$, $m_3$, $\dotsc$ are so selected that they approach zero as a
limit, it follows that $B$ is the limit point of the sequence $B_1$, $B_2$, $B_3$, $\dotsc$.

If a different sequence of divisors of $CD$, as $m_1'$, $m_2'$, $m_3'$, $\dotsc$, is used,
we obtain a sequence of points $B_1'$, $B_2'$, $B_3'$, $\dotsc$, likewise satisfying the
conditions of the problem.

Any two such sequences of points $B_1$, $B_2$, $B_3$, $\dotsc$ and $B_1'$, $B_2'$, $B_3'$, $\dotsc$
determined as above, are both \emph{increasing} and each is such that \emph{no} point
of it is to the right of \emph{every} point of the other.

Hence, by §~\Sref{420} any two such sequences have the same limit point~$B$.
}% [** PP: End of \small]

\Paragraph{423.}{Limit of a Sequence of Segments.} If $B_1$, $B_2$, $B_3$, $\dotsc$ is a
\index{Limit@\textbf{Limit}!of segments}%{\Pnum{423}}%
sequence of points on the segment $AB$ having the limit~$B$, then
the segment $AB$ is said to \emph{be the limit of the sequence of segments}
$AB_1$, $AB_2$, $AB_3$, $\dotsc$.

\NewFolio %% -----File: 205.png---Folio 197-------

\Section{INCOMMENSURABLE RATIOS}

\Paragraph{424.}{Ratio of Two Incommensurable Segments.} The ratio of
\index{Incommensurables@\textbf{Incommensurables}}%{\Pnum{424}}%
\index{Irrational@\textbf{Irrational}!ratio}%{\Pnum{424}}%
\index{Ratio@\textbf{Ratio}!incommensurable}%{\Pnum{424}}%
two incommensurable segments has not been explicitly defined,
but this may now be done in terms of the \emph{limit of a sequence}.

Consider two incommensurable segments $AB$ and $CD$. Let
$a_1$, $a_2$, $a_3$, $\dotsc$ be the lengths of segments each commensurable
with $CD$, forming a sequence whose limit is the segment $AB$,
and let $b$ be the length of the segment~$CD$.

Then $\dfrac{a_1}{b}$, $\dfrac{a_2}{b}$, $\dfrac{a_3}{b}$, $\dotsc$ is an increasing bounded sequence having
a limit which we call~$R$.

If $a_1'$, $a_2'$, $a_3'$, $\dotsc$ are the lengths of segments forming any
other sequence whose limit is $AB$, the sequence $\dfrac{a_1'}{b}$, $\dfrac{a_2'}{b}$, $\dfrac{a_3'}{b}$, $\dotsc$ is
another increasing bounded sequence with limit~$R'$.

By §~\Sref{420} we now know that $R = R'$.

This number $R$ is defined as \emph{the ratio of the incommensurable
segments} $AB$ and~$CD$.

\Paragraph{425.}{Ratio of Two Incommensurable Arcs.} The considerations
of §§~\Sref{422}, \Sref{423}, \Sref{424} apply to two arcs of equal circles.

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{205a}{0.6\textwidth}
\end{center}
\vspace*{-0.125in}

\begin{Remark}
Thus, in the figure, the ratio of the two incommensurable arcs $AB$
and $CD$ is the limit of the sequence $\dfrac{AB_1}{CD}$, $\dfrac{AB_2}{CD}$, $\dfrac{AB_3}{CD}$, $\dotsc$, and the ratio
of the incommensurable angles $AOB$ and $COD$ is the limit of the sequence
$\dfrac{\angle AOB_1}{\angle COD}$, $\dfrac{\angle AOB_2}{\angle COD}$, $\dfrac{\angle AOB_3}{\angle COD}$, $\dotsc$.
\end{Remark}

\NewFolio %% -----File: 206.png---Folio 198-------

\Section{APPLICATIONS OF LIMITS TO PLANE GEOMETRY}

\begin{Theorem}{426.}{IV.}
\index{Commensurable@\textbf{Commensurable}!segments}%{\Pnum{426}}%
A line parallel to the base of a
triangle, and meeting the other two sides, divides them
in the same ratio.
\end{Theorem}

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent\Objective{Given the $\triangle ABC$ with $DE \parallel BC$ and
cutting $AB$ and $AC$.}
{To prove that $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.}

\Proof: Consider the case when $AD$
and $AB$ are incommensurable.

Let $D_1$, $D_2$, $D_3$, $\dotsc$ be a sequence
of points on $AB$ whose limit is~$D$.
Through these points draw parallels
to $BC$, meeting $AC$ in $E_1$, $E_2$, $E_3$, $\dotsc$.
\end{minipage}%
% [** Illustration]
\hfill\Input{206a}{1.75in}

Then $E$ is the limit of the sequence of points $E_1$, $E_2$, $E_3$, $\dotsc$.

For suppose it is not, and that there is a point $K$ on $AE$
such that there is no point of $E_1$, $E_2$, $E_3$, $\dotsc$ between $K$ and~$E$.
Draw a line parallel to $BC$ through~$K$, meeting $AB$ in~$H$.

But there are points of the sequence $D_1$, $D_2$, $D_3$, \emph{between} $H$
and~$D$, and hence there must be points of the sequence $E_1$, $E_2$,
$E_3$, $\dotsc$ between $K$ and~$E$, which shows that $E$ is the limit of
the sequence $E_1$, $E_2$, $E_3$, $\dotsc$.

Then the sequence (1)~$\dfrac{AD_1}{AB}$, $\dfrac{AD_2}{AB}$, $\dfrac{AD_3}{AB}$, $\dotsc$ has a limit~$R$
which, by definition, is the ratio of the segments $AD$ and~$AB$.
\hfill §~\Sref{424}

Similarly, the sequence (2)~$\dfrac{AE_1}{AC}$, $\dfrac{AE_2}{AC}$, $\dfrac{AE_3}{AC}$, $\dotsc$ has a limit
$R'$ which is the ratio of $AE$ to~$AC$.

Now $\dfrac{AD_1}{AB} = \dfrac{AE_1}{AC}$, $\dfrac{AD_2}{AB} = \dfrac{AE_2}{AC}$, $\dfrac{AD_3}{AB} = \dfrac{AE_3}{AC}$, $\dotsc$ §~321, P.~G.
% [** PP: ``Plane Geometry''; period added]

Hence, the two sequences, (1) and~(2), are identical and have
the same limit (§~\Sref{419}). Hence $R = R'$.
\QED

\NewFolio %% -----File: 207.png---Folio 199-------

\Section{APPLICATIONS TO PLANE GEOMETRY}

\begin{Theorem}{427.}{V.}
\index{Commensurable@\textbf{Commensurable}!angles, arcs}%{\Pnum{427}}%
In the same circle or in equal
circles the ratio of two central angles is the same as the
ratio of their intercepted arcs.
\end{Theorem}

% [** Illustrations]
\vspace*{-0.125in}
\begin{center}
\Input{207a}{0.6\textwidth}
\end{center}
\vspace*{-0.125in}

\Proof: In case the arcs are commensurable the proof is
obvious. For, in that case, the common measure of the arcs
divides each into a number of equal parts, say $m$ and~$n$ respectively,
and these arcs subtend the same number of equal angles
at the center (§~\Sref{244}, Plane Geometry.) Hence the ratio of the
arcs is $m : n$ and the ratio of the angles is also $m : n$.

If the arcs $AB$ and $CD$ are not commensurable, let $AB_1$,
$AB_2$, $AB_3$, $\dotsc$ be a sequence of arcs whose limit is~$AB$, each
arc being commensurable with the arc~$CD$.

Then the sequence $\dfrac{AB_1}{CD}$, $\dfrac{AB_2}{CD}$, $\dfrac{AB_3}{CD}$, $\dotsc$ has a limit $R = \dfrac{AB}{CD}$.

Similarly the sequence $\dfrac{\angle AOB_1}{\angle COD}$, $\dfrac{\angle AOB_2}{\angle COD}$, $\dfrac{\angle AOB_3}{\angle COD}$, $\dotsc$ has
a limit $R' = \dfrac{\angle AOB}{\angle COD}$.

But $\dfrac{AB_1}{CD} = \dfrac{\angle AOB_1}{\angle COD}$, $\dfrac{AB_2}{CD} = \dfrac{\angle AOB_2}{\angle COD}$, $\dotsc$,
as shown above for the commensurable case.

Hence, these two sequences are identical and have the same
limit (§~\Sref{419}). Therefore it follows that $R = R'$.
\QED

\begin{Remark}
\textit{Note}.---The student should note the successive steps in the proofs of
§§~\Sref{426}, \Sref{427}: (a)~Definitions of the incommensurable ratios by means
of infinite sequences; (b)~Proof that these sequences are identical;
(c)~Conclusion that the limiting ratios are equal.
\end{Remark}

\NewFolio %% -----File: 208.png---Folio 200-------

\Paragraph{428.}{Products of Irrational Numbers.} If  $a_1$, $a_2$, $a_3$, $\dotsc$ is a
sequence with limit~$a$, then $ka_1$, $ka_2$, $ka_3$, $\dotsc$ is a sequence whose
limit is \emph{defined} as $ka$ where $k$ is any number.

If $a_1$, $a_2$, $a_3$, $\dotsc$ and $b_1$, $b_2$, $b_3$, $\dotsc$ are two sequences with limits
$a$ and~$b$ respectively, then $a_1b_1$, $a_2b_2$, $a_3b_3$, $\dotsc$ is a sequence whose
limit is \emph{defined} as~$ab$.

Similarly if $a_1$, $a_2$, $a_3$, $\dotsc$, $b_1$, $b_2$, $b_3$, $\dotsc$ and $c_1$, $c_2$, $c_3$, $\dotsc$ are
sequences with limits $a$, $b$, $c$, then $abc$ is defined as the limit
of the sequence $a_1b_1c_1$, $a_2b_2c_2$, $a_3b_3c_3$, $\dotsc$.

For a complete treatment it would be necessary to show that
these definitions of multiplication of irrational numbers are
consistent with the rest of arithmetic and also that these new
sequences are such as to determine definite limits. This is
possible but is beyond the scope of this book.

\Paragraph{429.}{Area of a Rectangle.} If the sides of a rectangle are incommensurable
with the unit segment, we \emph{define its area} as
follows:
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{429}}%

Let $a_1$, $a_2$, $a_3$, $\dotsc$ be a sequence of rational numbers whose
limit is the altitude~$a$; and let $b_1$, $b_2$, $b_3$, $\dotsc$ be a sequence whose
limit is the base~$b$.

Then the area of the rectangle is the limit of the sequence $a_1b_1$,
$a_2b_2$, $a_3b_3$, $\dotsc$.

% [** Illustration]
\begin{center}
\Input{208a}{0.4\textwidth}
\end{center}

But by definition the limit of $a_1b_1$, $a_2b_2$, $a_3b_3$, $\dotsc$ is the product
$ab$. Hence we have the theorem:

\begin{Theorem}{430.}{VI.}
\index{Area@\textbf{Area}!of a rectangle}%{\Pnum{430}}%
The area of a rectangle is the
product of its base and altitude.
\end{Theorem}

\NewFolio %% -----File: 209.png---Folio 201-------

\Section{APPLICATIONS TO PLANE GEOMETRY}

\Paragraph{431.}{Incommensurable Ratios Related to Two Circles.} In a
\index{Incommensurables@\textbf{Incommensurables}}%{\Pnum{431}}%
\index{Sequence@\textbf{Sequence}!approximating}%{\Pnum{431}}%
circle inscribe a sequence $P_1$, $P_2$, $P_3$, $\dotsc$ of regular polygons,
each having twice the number of sides of the one preceding it.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent Let the perimeters of these polygons be $p_1$, $p_2$, $p_3$, $\dotsc$ and
their areas $A_1$, $A_2$, $A_3$, $\dotsc$. Then the
length $c$ of the circle is defined to be
the limit of the sequence $p_1$, $p_2$, $p_3$, $\dotsc$
and its area $A$ is defined to be the
limit of $A_1$, $A_2$, $A_3$, $\dotsc$.

\Indent The sequence of polygons thus inscribed
is called an \emph{approximating
sequence of polygons}.
\end{minipage}%
% [** Illustration]
\hfil\Input{209a}{1.5in}\hfil

That these sequences are \emph{increasing}
and \emph{bounded} is obvious from the figure.

\begin{Theorem}{432.}{VII.}
The lengths of two circles are
in the same ratio as their radii, and their areas are in
the same ratio as the squares of their radii.
\end{Theorem}

\Proof: Let the radii of the two circles be $r$ and~$r'$. Denote
the ratio $r':r$ by~$k$. Then $r' = kr$.

Inscribe in one circle an approximating sequence of polygons
with perimeters $p_1$, $p_2$, $p_3$, $\dotsc$ and areas $A_1$, $A_2$, $A_3$, $\dotsc$. In the
other circle inscribe a sequence of similar polygons. By
§§~476, 477 (Plane Geometry), the perimeters of the latter are
$kp_1$, $kp_2$, $kp_3$, $\dotsc$ and their areas are $k^2A_1$, $k^2A_2$, $k^2A_3$, $\dotsc$.

By §~\Sref{428}, if the limit of $p_1$, $p_2$, $p_3$, $\dotsc$ is $c$ and the limit of
$A_1$, $A_2$, $A_3$, $\dotsc$ is $A$, then the limits of $kp_1$, $kp_2$, $kp_3$, $\dotsc$, and
$k^2A_1$, $k^2A_2$, $k^2A_3$, $\dotsc$ are~$kc$ and~$k^2A$, respectively.

That is, the ratio of the lengths of the circles is $\dfrac{kc}{c} = k= \dfrac{r'}{r}$
and the ratio of their areas is $\dfrac{k^2A}{A} = k^2 = \dfrac{r'^2}{r^2}$.

\NewFolio %% -----File: 210.png---Folio 202-------

\Section{APPLICATIONS OF LIMITS TO SOLID GEOMETRY}

\Paragraph{433.}{Incommensurable Solids.} The areas of curved surfaces,
\index{Incommensurables@\textbf{Incommensurables}}%{\Pnum{433}}%
the volumes of cones, cylinders, and spheres, and even of rectangular
solids involve \emph{incommensurable quantities}. We have
already treated these cases informally and have devised means
for computing them \emph{approximately}.

We now give a formal treatment for logical completeness.

\Paragraph{434.}{Rectangular Parallelopiped.} If the three concurrent
\index{Volume@\textbf{Volume}!as a limit of a sequence}%{\Pnum{434}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{434}}%
edges $a$, $b$, $c$ of a rectangular parallelopiped are incommensurable
with the unit segment, the
volume inclosed is defined as follows:

Let $a_1$, $a_2$, $a_3$, $\dotsc$ be a sequence of
rational numbers whose limit is the
side~$a$. Let $b_1$, $b_2$, $b_3$, $\dotsc$ and $c_1$, $c_2$, $c_3$, $\dotsc$
be similar sequences whose limits
are respectively the dimensions $b$ and~$c$.

\noindent\begin{minipage}[b]{\textwidth-1.75in}
\Indent Then the volume is the limit of the
sequence $a_1b_1c_1$, $a_2b_2c_2$, $a_3b_3c_3$, $\dotsc$.

But the limit of this sequence is by
definition the product of the limits of the three sequences
$a_1$, $a_2$, $a_3$, $\dotsc$, $b_1$, $b_2$, $b_3$, $\dotsc$, $c_1$, $c_2$, $c_3$, $\dotsc$, or $abc$.
\end{minipage}%
% [** Illustration]
\hfill\smash{\raisebox{-0.25in}{\Input{210a}{1.625in}}}

Hence, we have the theorem:

\begin{Theorem}{435.}{VIII.}
The volume of a rectangular
parallelopiped is equal to the product of its three
concurrent edges.
\end{Theorem}

\begin{Exercises}

\item[1.] Give in decimals four terms of a sequence which approximates
the volume of a rectangular parallelopiped whose
dimensions are $2$, $4$, $\sqrt{5}$.

\item[2.] Give in decimals four terms of a sequence which approximates
the circumference of a circle whose radius is $10$~in.;
also of a sequence which approximates the area.

\end{Exercises}

\NewFolio %% -----File: 211.png---Folio 203-------

\Section{APPLICATION OF LIMITS TO THE PYRAMID}

\Paragraph{436.}{Triangular Pyramid.} In a triangular pyramid inscribe
a set of prisms in a manner similar to that shown in the figure
of §~\Sref{243}.
\index{Circumscribed@\textbf{Circumscribed}!prism}%{\Pnum{436}}%
\index{Inscribed@\textbf{Inscribed}!prism}%{\Pnum{436}}%
\index{Prism@\textbf{Prism}!inscribed}%{\Pnum{436}}%
\index{Volume@\textbf{Volume}!as a limit of a sequence}%{\Pnum{436}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{436}}%

Call $V_1$ the sum of the volumes of these inscribed prisms.

Using as altitudes one half the altitudes of the first set of
prisms, inscribe a second set the sum of whose volumes is~$V_2$.
Continuing in this manner, we obtain a sequence of sets of
prisms with volumes $V_1$, $V_2$, $V_3$, $\dotsc$. The limit $V$ of this sequence
we define as \textbf{the volume of the pyramid}.

\begin{Remark}
Circumscribed prisms may also be used for defining the volume of a
pyramid, in which, case we get a \emph{decreasing} sequence of volumes
$U_1$, $U_2$, $U_3$, $\dotsc$, with limit~$U$. That $U = V$ follows from the fact that
the sum of the volumes of the circumscribed prisms exceeds that of the
inscribed prisms by the volume of the lowest circumscribed prism, and
this may be made as small as we please.
\end{Remark}

\begin{Theorem}{437.}{IX.}
Two pyramids with the same
altitudes and equivalent bases have equal volumes.
\end{Theorem}

\Proof: Call a sequence of inscribed prisms in one pyramid
$V_1$, $V_2$, $V_3$, $\dotsc$, and in the other $V'_1$, $V'_2$, $V'_3$, $\dotsc$. Since corresponding
sets of these prisms have equal volumes (§~\Sref{200}), that
is, $V_1 = V'_1$, $V_2 = V'_2$, $V_3 = V'_3$, % [** PP: Added comma]
$\dotsc$, the theorem follows from
§~\Sref{419}.

\Paragraph{438.}{Convex Closed Curves.} We assume that in a \emph{convex
closed curve} (§~\Sref{202}) it is possible to inscribe a sequence of
\index{Curves@\textbf{Curves}!convex, closed}%{\Pnum{438}}%
polygons $P_1$, $P_2$, $P_3$, $\dotsc$, having perimeters $p_1$, $p_2$, $p_3$, $\dotsc$ and
areas $A_1$, $A_2$, $A_3$, $\dotsc$ with limits $p$ and $A$ respectively, and to
circumscribe a sequence of polygons $P'_1$, $P'_2$, $P'_3$, $\dotsc$, having
perimeters $p'_1$, $p'_2$, $p'_3$, $\dotsc$ and areas $A'_1$, $A'_2$, $A'_34$, $\dotsc$ with limits
$p'$ and $A'$ respectively, such that $p = p'$ and $A = A'$.

These limits $p$ and $A$ we now define as \textbf{the perimeter and the
area respectively of the curve}.

\NewFolio %% -----File: 212.png---Folio 204-------

\Section{APPLICATION OF LIMITS TO THE CYLINDER}

\Paragraph{439.}{The Cylinder.} Given any cylinder with a convex right
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{439}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{439}}%
cross-section and an element~$e$. In this cross-section inscribe
a sequence $P_1$, $P_2$, $P_3$, $\dotsc$ of polygons, as in §~\Sref{438}, with perimeters
$p_1$, $p_2$, $p_3$, $\dotsc$ and areas $A_1$, $A_2$, $A_3$, $\dotsc$, thus defining the
perimeter $p$ and the area $A$ of the cross-section.

Consider a set of prisms inscribed in this cylinder, of which
$P_1$, $P_2$, $P_3$, $\dotsc$ are right cross-sections.

Then the areas and the volumes of these prisms are respectively
$p_1e$, $p_2e$, $p_3e$, $\dotsc$ and $A_1e$, $A_2e$, $A_3e$, $\dotsc$.

\textbf{The lateral area and the volume of the cylinder are now defined
as the limits of these sequences}.\quad But by §~\Sref{428} these limits are
equal respectively to $pe$ and~$Ae$.

Hence, we have the theorem:

\begin{Theorem}{440.}{X.}
The lateral area of a cylinder is
the product of an element, and the perimeter of a
right section and its volume is the product of an element
and the area of a right section.
\end{Theorem}

\begin{Exercises}

\item[1.] Prove as above that the volume of any convex cylinder
is equal to the product of its altitude and the area of its base.

\item[2.] Prove that the lateral area of a right circular cone is
equal to half the product of the slant height and the perimeter
of its base.

\item[3.] Prove that the volume of any convex cone is equal to
one third the product of its altitude and the area of its base.

\begin{Suggestion}
The treatment required in these exercises is a very close
paraphrase of the definitions and proof given in §~\Sref{439}. Observe that we
cannot begin to make a proof until we have \emph{defined} the subject matter
of the theorem. That is, we must first define \textbf{the areas and volumes in
question}.
\end{Suggestion}
\end{Exercises}

\NewFolio %% -----File: 213.png---Folio 205-------

\Section{APPLICATION OF LIMITS TO THE SPHERE}

\Paragraph{441.}{The Sphere.} Through two points $P$ and $Q$ of a sphere
pass a great circle forming a hemisphere with center~$O$.
\index{Volume@\textbf{Volume}!as a limit of a sequence}%{\Pnum{441}}%

% [** Illustrations]
\begin{center}
\Input{213ab}{\textwidth}
\end{center}

Divide $OA$, the radius perpendicular to the plane of the
great circle, into the equal parts $OC$, $CB$, and $BA$. Through
$C$ and $B$ pass planes parallel to the plane of $POQ$, meeting
the sphere in points $D$ and~$E$, respectively.

Construct right circular cylinders with axes $OC$ and $CB$
and radii $CD$ and $BE$. Denote by $V_1$ the sum of the volumes
of these cylinders.

Now divide the radius $OA$ into six equal parts and construct
five cylinders in the same manner as above. Let the sum of
these volumes be $V_2$.

Continuing in this manner, each time dividing $OA$ into
twice as many equal parts as in the preceding, we obtain a
sequence of sets of cylinders and a corresponding sequence
$V_1$, $V_2$, $V_3$, $\dotsc$ of volumes.

We now define the volume of the hemisphere as the limit of
the sequence $V_1$, $V_2$, $V_3$, $\dotsc$.

\Paragraph{442.}{} Construct a right circular cylinder with its base in the
plane of $POQ$ and with radius and altitude both equal to~$OA$.

Denote by $F$ the figure formed by the lower base of the
cylinder, its lateral surface and the lateral surface of the cone
whose base is the upper base of the cylinder and whose vertex
is at $O'$. (See the right-hand figure.)

\NewFolio %% -----File: 214.png---Folio 206-------

Draw segments $O'M$ and $O'N$. Let the planes through $C$
and $B$ cut $O'A'$ in $C'$ and $B'$ and $O'M$ in $H$ and $K$.

Now form the cylinder $O'C'H$ whose axis is $O'C'$ and whose
radius is $C'H$. Likewise form the cylinder $C'B'K$.

Let $V'_1$ denote the sum of the volumes of $O'C'D'$ and $C'B'E'$
minus the sum of the volumes of $O'C'H$ and $C'B'K$.

In a similar manner, using the planes which divide $OA$ and
hence $O'A'$ into six equal parts, we form another set of five
cylinders, the sum of whose volumes minus that of the smaller
inside cylinders we denote by $V'_2$.

Continuing in this manner, we obtain a sequence of volumes
$V'_1$, $V'_2$, $V'_3$, $\dotsc$ whose limit $V'$ we define as \textbf{the volume of the
given figure}~$F$.

We now prove that $V_1 = V'_1$, $V_2 = V'_2$, $\dotsc$.

Denote $OA$ by $r$, and note that $O'B' = B'K$.

% [** PP: Re-breaking next line]
(1)~$\text{Vol.\ } C'B'E' - \text{Vol.\ } C'B'K
  = \pi C'B'(\overline{B'E'}^2 - \overline{B'K}^2)$ \\
\indent$\phantom{\text{(1)~Vol.\ } C'B'E' - \text{Vol.\ } C'B'K}
  = \pi C'B'(r^2 - \overline{O'B'}^2)$.

% [** PP: Re-breaking]
(2)~$\text{Vol.\ } CBE
  = \pi CB \cdot \overline{BE}^2
  = \pi CB(r^2 - \overline{OB}^2)$, since \\
\indent$\overline{BE}^2 = r^2 - \overline{OB}^2$. But $OB = O'B'$ and $CB = C'B'$.

Hence, $\text{Vol.\ } CBE = \pi C'B'(r^2 - \overline{O'B'}^2)$.

Hence, $\text{Vol.\ } CBE = \text{Vol.\ } C'B'E' - \text{Vol.\ } C'B'K$.

Similarly we show that
\[
\text{Vol.\ } OCD = \text{Vol.\ } O'C'D' - \text{Vol.\ } O'C'H.
\]

Hence, $V_1 = V'_1$. In like manner $V_2 = V'_2$, $V_3 = V'_3$, $\dotsc$.

Hence, $V = V'$, since they are the limits of the same
sequences.

But the volume of the cylinder $O'A'M$ is $\pi r^3$ and of the cone
whose volume was subtracted, $\tfrac{1}{3} \pi r^3$. That is, the volume of
$F$ is $\tfrac{2}{3} \pi r^3$, and hence that of the hemisphere is $\tfrac{2}{3} \pi r^3$. % [** PP: Changed exponent to 3]

Hence, we have the theorem:

\begin{Theorem}{443.}{XI.}
\index{Sphere@\textbf{Sphere}!volume of}%{\Pnum{443}}%
\index{Volume@\textbf{Volume}!of a sphere}%{\Pnum{443}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{443}}%
The volume of the sphere is $\tfrac{4}{3} \pi r^3$.
\end{Theorem}

\NewFolio %% -----File: 215.png---Folio 207-------

Note that the above proof consists essentially in showing
that the area of the circle $BE$ is equal to that of the ring
between the circles $B'E'$ and $B'K$, and that the area of the
circle $CD$ is equal to that of the ring between $C'D'$ and $C'H$
and so on.

Indeed, this theorem and also that of §~\Sref{437} are special cases
of what is known as \textbf{Cavalieri's Theorem}.
\Page{202}%

\begin{Theorem}{444.}{XII.}
\index{Cavalieri@\textbf{Cavalieri's theorem}}%{\Pnum{444}}%
If two solid figures  are regarded
as resting on the same plane $b$, and if in every
plane parallel to $b$ the sections of the two figures have
equal areas, the figures have equal volumes.
\end{Theorem}

The proof of this general theorem is more difficult than any
thus far given, inasmuch as it involves sequences which \emph{oscillate};
that is, which are neither constantly increasing nor constantly
decreasing.

\Paragraph{445.}{The Area of the Sphere.} About a sphere of radius $r$
\index{Area@\textbf{Area}!of a sphere}%{\Pnum{445}}%
\index{Sphere@\textbf{Sphere}!area of}%{\Pnum{445}}%
\index{Measurement@\textbf{Measurement}!of surfaces}%{\Pnum{445}}%
\index{Measurement@\textbf{Measurement}!of volumes}%{\Pnum{445}}%
construct a sequence of circumscribed polyhedrons such that
the largest face in each polyhedron becomes as small as we
please when we proceed along the sequence. Let $s_1$, $s_2$, $s_3$, $\dotsc$
be the total surfaces of these polyhedrons. This forms a decreasing
sequence with limit $S$ which we define as \textbf{the surface
of the sphere}.

The volumes of these polyhedrons will be $\tfrac{1}{3}rs_1$, $\tfrac{1}{3}rs_2$, $\tfrac{1}{3}rs_3$, $\dotsc$.
Then the volume $V$ of the sphere is defined\footnote{This definition can be shown to be consistent with that of §~\Sref{441}.}
as \textbf{the limit
of this sequence of volumes}.

\settowidth{\TmpLength}{Hence, by §~\Sref{428}, }%
Hence, by §~\Sref{428}, $V = \tfrac{1}{3}rS$. But by §~\Sref{443}, $V = \tfrac{4}{3} \pi r^3$.

\makebox[\TmpLength][l]{Then}$S = \tfrac{3}{r} \cdot \tfrac{4}{3} \pi r^3 = 4 \pi r^2$.

Hence, we have the theorem:

\begin{Theorem}{446.}{XIII.}
The area of the sphere is $4 \pi r^2$.
\end{Theorem}

\NewFolio %% -----File: 216.png---Folio 208-------

\Section{EXERCISES ON LIMITS}

\begin{Itemize}

\item[1.] In addition to those which are found in the text give
other examples of infinite sequences which do not determine
definite numbers.

\item[2.] Give two increasing sequences which determine the
number~$3$. Show that the theorem of §~\Sref{420} applies and proves
that these sequences determine the same number.

\item[3.] Give two decreasing sequences each of which determines
the number~$5$. Apply §~\Sref{421} to show that these sequences
determine the same number.

\item[4.] State fully the relation between a bounded increasing
sequence and the number determined by it. State also the
relations between a bounded decreasing sequence and the number
determined by it.

\item[5.] State fully what is meant by ``a limit of a sequence''
both for increasing and decreasing sequences.

\item[6.] Given two incommensurable segments $AB$ and $CD$.
Lay off on the line $AB$ a \emph{decreasing} sequence of segments,
each of which is commensurable with $CD$, such that the limit
of the sequence is the segment~$AB$.

\item[7.] If $a_1$, $a_2$, $a_3$, $\dotsc$ is an increasing sequence defining the
number~$4$, prove that $3a_1$, $3a_2$, $3a_3$, $\dotsc$ defines the number
$3 × 4 = 12$.

\item[8.] If $a_1$, $a_2$, $a_3$, and $b_1$, $b_2$, $b_3$, $\dotsc$ are increasing sequences
defining the numbers $3$ and~$5$, show that the sequence $a_1b_1$,
$a_2b_2$, $a_3b_3$, $\dotsc$ defines the number~$15$.

\item[9.] Describe in more detail the meaning of the ratio of two
incommensurable arcs as indicated in §~\Sref{425}. Show that this
ratio is independent of the sequence of units of measurement
used, so long as the limit of this sequence is zero.

\item[10.] Treat the ratio of two incommensurable angles in a
manner similar to the treatment of arcs in the preceding
exercise.

\end{Itemize}

\NewFolio %% -----File: 217.png---Folio 209-------
% [Blank Page]

\NewFolio %% -----File: 218.png---Folio 210-------

%%%%%%%%%%%%%%%%%%%%%%%%%%%%% BACK MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%%%

\backmatter

%% Finish off the Table of Contents: Index, and Bio Sketch entries

% ad hoc hack to reduce number of overfull hboxes
\addtolength{\textwidth}{10pt}
%\addtolength{\textheight}{-24pt}

\fancyhead[C]{\textit{INDEX}}
\phantomsection
\addcontentsline{toc}{chapter}{\texorpdfstring{INDEX}{Index}}

% not \addcontentsline; no number
\addtocontents{toc}{\protect\contentsline{chapter}{PORTRAITS AND BIOGRAPHICAL SKETCHES}{}{}}

% [** PP: ``Frontispiece'' would overlap dot leaders]
% To prevent this, write the code below into the toc file:
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% \settowidth{\FPlen}{\small\textit{Frontispiece}}
% \makeatletter
% \renewcommand{\@pnumwidth}{\FPlen}
% \makeatother

\addtocontents{toc}{\protect\newlength{\protect\FPlen}}
\addtocontents{toc}{\protect\settowidth{\protect\FPlen}%
  {\protect\small\protect\textit{Frontispiece}}}
\addtocontents{toc}{\protect\makeatletter}
\addtocontents{toc}{\protect\renewcommand{\protect\@pnumwidth}{\protect\FPlen}}
\addtocontents{toc}{\protect\makeatother}

% Set the ToC entry...
% @@@@ If linktocpage=true, move four \hyperrefs onto page numbers
\phantomsection
\addtocontents{toc}{\protect\contentsline{section}{\hyperref[page:Cavalieri]{Cavalieri}}%
{\protect\llap{\protect\small\protect\textit{Frontispiece}}}{}}

% And restore \@pnumwidth
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\addtocontents{toc}{\protect\makeatletter}
\addtocontents{toc}{\protect\renewcommand{\protect\@pnumwidth}{\protect\FPlen}}
\addtocontents{toc}{\protect\makeatother}

\phantomsection
\addtocontents{toc}{\protect\contentsline{section}{\hyperref[page:Thales]{Thales}}%
{\pageref*{page:Thales}}{}}

\phantomsection
\addtocontents{toc}{\protect\contentsline{section}{\hyperref[page:Archimedes]{Archimedes}}%
{\pageref*{page:Archimedes}}{}}

\phantomsection
\addtocontents{toc}{\protect\contentsline{section}{\hyperref[page:Legendre]{Legendre}}%
{\pageref*{page:Legendre}}{}}

% Oh, and incidentally, print the index...
{
\printindex
}

\cleardoublepage

%%%%%%%%%%%%%%%%%%%%%%%%% TRANSCRIBER'S NOTE %%%%%%%%%%%%%%%%%%%%%%%%%

\pdfbookmark[0]{Back Matter}{Back Matter}

\phantomsection
\pdfbookmark[1]{Transcriber's Notes}{Transcriber's Notes}
\fancyhead[C]{\textit{TRANSCRIBER'S NOTES}}

\Section{Transcriber's Notes}

The spellings \emph{inclose} and \emph{parallelopiped} have been
retained. Minor typographical corrections and regularizations of
spelling have been made without comment.
\medskip

\textbf{\hyperref[err019a]{Page~\pageref{err019a}}} In the
third (rightmost) figure after Theorem~73, the labels $\ell_1$
and~$\ell_2$ are exchanged relative to their use in the subsequent
exposition.
\Note{019a}
\medskip

The book contains at least five potentially ambiguous assertions. Some
are complicated to resolve completely.

\textbf{\hyperref[err058a]{Page~\pageref{err058a}}} (\emph{How many
vertices and edges has a hexahedron}?) Combinatorially there are
multiple types of hexahedron, including but not limited to (i)~a
parallelepiped (8~vertices and 12~edges), (ii)~a square-based pyramid
with one base corner shaved off (7~vertices and 11~edges), (iii)~a
pentagonal pyramid (6~vertices and 10~edges), and (iv)~a compound of
two regular tetrahedra placed base-to-base (5~vertices and 9~edges).
\Note{058a}

\textbf{\hyperref[err078a]{Page~\pageref{err078a}}} (\emph{A
closed plane curve is \emph{convex} if it can be cut by a straight
line in not more than two points}.) Presumably this was to be
interpreted as ``\ldots if it \textbf{cannot} be cut by a straight
line in \textbf{more} than two points.'' However, even the latter is
inconsistent with usage throughout the book, since some polygons are
convex while no polygon satisfies this condition.
\Note{078a}

\textbf{\hyperref[err082a]{Page~\pageref{err082a}}}: Approximating a
curved surface by planar patches is more subtle than the analogous
approximation of a curve by an inscribed sequence of segments. For
example, bounded polyhedra inscribed in a right circular cylinder may
have arbitrarily large area.
\Note{082a}

\textbf{\hyperref[err152a]{Page~\pageref{err152a}}}: The construction
of a sequence of polyhedra circumscribed about a sphere depends on the
points of tangency chosen at each stage, and these points cannot be
chosen arbitrarily.
\Note{152a}

\textbf{\hyperref[err197a]{Page~\pageref{err197a}}}: The ``definition''
of~$\sqrt{2}$ as the limit of its sequence of finite decimal
truncations is circular since it requires the first $n$~decimal places
to be known in order to generate $n$~terms of the sequence.
\Note{197a}

\cleardoublepage

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\phantomsection
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\fancyhead[C]{\textit{LICENSING}}

\begin{PGtext}
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Applications (Revised edition), by H. E. Slaught and N. J. Lennes

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