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Font size the area wUl be the greatest^ when the given sides are so
disposedf that the figure moAf be inscridbd in a semicir-
cle, of which the undetermined side is the diaimeter*

If the sides AB, BC, CD, DE, (Fig. 28.) be given, and if
their position be such that the area, included l^tween these
and another side whose length is not determined, is a maxi-
mum ; tfie figure may be inscribed in a semicircle, of which
the undetermined side AE is the diameter.

Draw the lines AD, AC, EB, EC. By varying; the asij^fe
at D, the triangle ADE may be enlarged or diminished, with-
out affecting the area of the other parts of the figure. The
whole area, therefore, cannot be a m^aximum^ unless -this
triangle be a mairmiÂ«m, while the sides AD and ED aie
given. But if the triangle ADE be a maximum,^ under these
therefore the point D is in the circumference of a circle, of
which AE is the diameter. (Euc. 31, 3^ In the same manner
it may be proved, that the angles ACE and ABE are right
angles, and therefore that the points. C and B ore in the cir-
cumference of the same circle.

The term polygon is used in this section to include irir
angles J and four-sided figures, as well as othea: riglit4ined
figures.

82. The area of a polygon, inscribed in a semicifcle, in the
manner stated above, will not be altered by varyis^lhe ^naisr
^f the given sides.

The sides AB, BC, CD, DE, (Fig. 28.) are the dhÂ»rdjs of
so many arcs. The sum of these mrcs, in whatever ordeo:
they are arranged, will evidently be equal to the senieirciuii-
ference. And the segm^ents between the gMFen odes fludliie*

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^ EM>raUMETRY. ^

arcs will be the same, in whatever part of the circle they are
situated. But the area of the polj^n is equal to the area <rf
the semicircle, diminished by Uie sum of these segments.

83, If a polygon, of" which all the sides except one are
given, be inscnted in a semicircle whose diameter is the un-
determined side; a polygon having the same given sides,
cannot be inscribed in any other semicircle which is either
greater or less than this, and whose diameter is the undeter-
mined side.

The given sides AB, EC, CD, DE, (Fig. 28.) are the
chords of arcs whose sum is 180 degrees. But in a larger
circle, eadi would be the chcÂ»rd of a less number of degrees,
and therefore the sum of the arcs would be less than 18(P:
and in a smaller circle, each would be the chord of a greater
number of degrees, and the sum of the arcs would be greater
than 180Â°.

PROPOSITION IV.

84 A polygon inscribed in a circle htzs a greater
arefij than any polygon of equal perimeter^ and the same ^
number ofsidesj which cannot be inscribed in a circle.

If in the circle ACHF, (Fig. 30.) there be inscribed a poly-
gon ABCDEFG ; and if another polygon abcdefg (Fig. 31.)
be formed of sides which are the same m numba* and length,
but which are so disposed, that the figure cannot be inscribed
in a circle ; the area of the former polygon is greater than
that of the latter.

Draw the diameter AH, and the chords DH and EH.
Upon de make the triangle deh equal and similar to DEH,
and join ah. ' The line ah divides the figure abcdhefg into
two parts, of which one at least cannot, by supposition, be
inscribed in a semicircle of which the diameter is AH, nor
in any other semicircle of which the diameter is the undeter-
mined side. (Art. 83.) It is therefore less than the corres-
ponding part of the figure ABCDHEFG. (Art. 81.) And
the other part of abcdhefg is not greater than the correspon-
ding part of ABCDHEFG. Therefore, the whole figure
ABCDHEFG is greater than the whole figure abcdhefg.
If firom these there be taken the equal triangles DEH and
de&i th^re will remain the polygon ABCDEFG greater than
the polygon abedefg.

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I80PERIMETRT. gj^

86. A polygon of which all the sides are ffiven in nnmber
and length, can not be inscribed in circles of different diame-
ters. (Art. 83.) And the area of the polygon will not be alter-
ed, by changing the order of the sides. (Art. 82.)

PROPOSITION V.

86. When a polygon has a greater area than any other ^
of the same number of sideSj and of equal perimeter y the
sides are e^ual.

The polygon ABCDP (Fig. 29.) cannot be a maximum^
among all polygons of the same number of sides, and of equal
perimeters, unless it be equilateral. For if any two of the
sides, as CD and FD, are unequal, let CH and FH be equal,
and their sum the same as the sum of CD and FD. The
isosceles triangle CHF is greater than the scalene triangle
CDF (Art. 79.); and therefore the polygon ABCHF is greater
than the polygon ABCDF ; so that the latter is not a max-
vm/wm.

PROPOSITION VI.

87. A REGULAR POLYGON hos a greater area than emy
other polygon of equal perimeter, and of the same number
of sides.

For, by the preceding article, the polygon which is a max-
imum amon^ others of equal perimeters, and the same num-
ber of sides, IS equilateral, and by art. 84, it itoay be inscribed
in a circle. But if a polygon inscribed in a circle is equilat-
eral, as ABDFGH (Fig. 7.) it is also equiangular. For the
sides of the polygon are the bases of so many isosceles trian-
gles, whose common vertex is the center C. The angles at
these bases are all equal ; and two of them, as AllC and
GHC, are equal to AHG one of the angles of the polygon.
The polygon, then, being equiangular, as well as equilateral,
is a regular polygon. [Axi. 1. Def 2.)

Thus an equilateral triangle has a greater area, than any
other triangle of equal perimeter. And a square has a greater
area, than any other four-sided figure of equal perimeter.

9 X i

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Qg laOPERniETRT.

Cor. A regular* polygon has a less perimeter than any other
polygon of equal area, and the same number of sides.

For if, with a given perimeter, the re^lar polygon is
greater than one which is not regular ; it is evident the pe-
rimeter of the former must be diminished, to make its area
equal to that of the latter.

PROPOSITION VII.

S8. ]^ a polygon be described aboitt a circle, the
areas <^the two figures are as their perimeters.

Let ST (Fig. 32.) be one of the sides of a polygon, either
regular or not, which is described about the circle LNR.
Join OS and OT, and to the point of contact M draw the
radius OM, which will be perpendicular to ST. (Euc. 18, 3.^
The triangle OST is equal to half the base ST multiplied
into the radius OM. (Art. 8.) And if lines be drawn, in the
same manner, from the center of the circle, to the extremities
of the several sides of the circumscribed polygon, each of the
triangfles thus formed will be equal to half its base multiphed
into the radius of the circle. Therefore the area of the whole
polygon is equal to half its perimeter multiplied into the
radius : and the area of the circle is equal to half its circum-
ference multiplied into tiie radius. (Art. 30.) So that the two
areas are to each other as their perimeters.

Cor. i. If different polygons are described about the same
circle, their areas are to each other as their perimeters. For
tiie area of each is equal to half its perimeter, multiplied into
the radius of the inscribed circle.

Cor. 2. The tangent of an arc is alwa3rs greater than the
arc itself The triangle OMT (Fig. 32.) is to OMN, as MT
to MN. But OMT is greater than OMN, because the former
includes the latter. Therefore, the tangent MT is greater
than the arc MN.

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ISOPBRIMETRT. QJ

PROPOSITION VIII.

89. A CIRCLE has a greater area than any polygon of
equal peri/meter.

If a circle and a regular polygon have the same center, and
equal perimeters ; each of the sides of the polygon must fall
partly toithin the circle. For the area o/ a circumscribing
polygon is greater than the area of the circle, as the one in-
cludes the other : and therefore, by the preceding article, the
perimeter of the former is greater than that of the latter.

Let AD thei^ (Fig. 32.) be one side of a regular polygon,
whose perimeter is equal to the circumference of the circle
RLN. As this falls partly within the circle, the perpendicu-
lar OP is less than the radius OR. But the area of the poly-
gon is equal to half its perimeter multiplied into this perpen-
dicular (Art. 16.) ; and the area of the circle is equal to half
its circumference multiplied into the radius. (Art. 30.) The
circle then is greater than the given regular polygon ; and
therefore greater than any other polygon of equal perimeter.
(Art. 87.)

Cor. 1. A circle has a less perimeter^ than any polygon of
equal area.

Cor. 2. Among regular polygons of a given perimeter, that
which has the greatest number of sides, has also the great-
est area. For the greater the number of sides, the more
nearly does the perimeter of the polygon approach to a coin-
cidence with the circumference of a circle.*

PROPOSITION IX.

90. A right prism whose bases are regular polygons,
has a less surface than any other right prism of the same
solidity, the same altitude, and the same number of sides.

If the altitude of a prism is given, the area of the base is
as the solidity (Art. 43.) ; and if the number of sides is also
given, the perimeter is a minimum, when the base is a regular

* For a rigorous demonstration of this, see Legendre's Geometry, Appendix to
Book IT.

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^ ISOPSBIMETEY.

polygon. (Art 87. Cor.) But the lateral surface is as the per-
imeter. (Art. 47.) Of two right prisms, then, which have the
same altitude, the same solidity, and the same number of
sides, that whose bases are regular polygons has the least
lateral surface, while the areas of the ends are equal.

Cor. A right prism whose bases are regular polygons has
a greater solidity^ than any other right prism of me same
surface, the same altitude, and the same number of sides.

PROPOSITION X.

91. A right cylinder ha^ a less surface, than any right
prism of the same altitude and solidity.

For if the prism and cyhnder have the same altitude and
solidity, the areas of their bases are equal. (Art. 64.) But the
perimeter of the cylinder is less, than that of the prism (Art.
89. Cor. 1.) ; and therefore its lateral surface is less, while the
areas of the ends are equal.

Cor. A right cylinder has a greater solidity, than any
right prism of the same Â«dtitude and surface.

PROPOSITION XI.

92. A CUBE has a less surface than any other right parol-
Idopiped of the saine solidity.

A parallelepiped is a prism, any one of whose faces may be
considered a base. (Art. 41. Def. I. and V.) If these are not
all squares, let one which is not a square be taken for a base.
The perimeter of this may be diminished, without altering
its area (Art. 87. Cor.) ; and therefore the surface of the solid
may be diminished, without altering its altitude or solidity.

SArt. 43, 47.) The same may be proved of each of the other
aces which are not squares. The surface is l^bcrefore a
minimum, when all the faces are squares, that is, when tha
solid is a cube.

Cor. A cube has a greater solidity than any other right
parallelepiped pf the same surfece.

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nOPERflllETRY.

PROPOSITION XII.

69

93. A CUBE has a greater solidity, than any other right
paraUelopipedj the sum of whose length, breadth, and depth
is equal to the sum of the corresponding dimensions of the
cube.

The solidity is equal to the product of the length, breadth,
and depth. If the length and breadth are unequal, the soUd-
ity may be increased, without altering the sum of the three
dimensions. For the product of two factors whose mim is

g' ven, is tl^ greatest when the factors are equal. (Euc. 27. 6.)
L the isame manner, if the breadth and depth are unequal,
the solidity may be increased, without altering the sum of
the three dimensions. Therefore, the solid can not be a
maximum, unless its length, breadth, and depth are equsJL

PROPOSITION XIII.
94. Iff' a PRISM BE DESCRIBED ABOUT A CYLINDER, the

capacities of the two solids are as their surfaces.

The capacities of the solids are as the areas of their bases,
that is, as the perimeters of their bases. (Art. 88.) But the
lateral surfaces are also as the perimeters of the bases. There-
fore the whole surfaces are as the solidities.

Ck)r. The capacities of different prisms, described about the
isame right cylinder, are to each other as their surfaces.

PROPOSITION XIV.

95. A right cylinder whose height is EauAL to the
DIAMETER OF ITS BASE hos a greater solidity than any
other right cylinder of equal surface.

Let C be a right cylinder whose height is equal to the di-
ameter of its base ; and C^ another right â€˘cylinder having the
same surface, but a different altitude. If a square prism P be.
described about the former, it will be a cuhe. But a square
prism P' described about the latter will not be a cube.

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70 laOPERIBfETRT.

Then the surfaces of C and P are as their bases (Art. 47.
and 88.) ; which are as the bases of C and P', (Sup. Euc. ^^
1.) ; so that,

aurfOisurfV: ibaseGlbaseP: ibaseC' ibasePf i isurfC'isurfP'

But the surface )ff C is, by supposition, equal to the sur&ce
of C Therefore, ( Alg. 395.) the surface of P is equal to the
surface of P'. Ana by the preceding article,

solidP : solidC : : surfP : surfC : : surfP' : 0urfC' : : soltdF' : solidC'

But the solidity of P is greater than that of P'. (Art 92.
Ck)r.) Therefore, the solidity of C is greater tlMui that of C

Schol. A right cylinder whose height is equal to the diam-
eter of its base, is that which circu^nscribes a sphere. It ia
also called Archimede^ cylinder ; as he discovered the ratio
of a sphere to its circumscribing cylinder ; and these are the
figures which were put upon his tomb.

Cor. Archimedes' cylinder has a less surface^ than any
other right cylinder of the same capacity.

PROPOSITION XV.

%. ijf a SPHERE BE CIRCUMSCRIBED by a soUd bouuded
by plane surfaces ; the capacities of the two solids are as
their surfaces.

If planes be supposed to be drawn from the center of the
sphere, to e^ch of the edges of the circumscribinff sohd, they
will divide it into as many pyramids as the solid has &ces.
The base of each pyramid will be one of the faces ; and the
height will be the radius of the sphere. The capacity of the
pyramid will be equal, therefore, to its base multiplied into \
of the radius (Art. 48.) ; and the capacity of the whole cir-
cumscribing solid, must be equal to its whole surface multi-
pUed into \ of the radius. But the capacity of the sphere is
also equal to its surface multiplied into \ of its radius. (Art.
70.)

Cor. The capacities of different solids circumscribing the
same sphere, are as their surfaces.

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I90PEB0ICETRY. 71

PROPOSITION XVI.

97. A SPHERE has a greater solidity than any regular
fclyedron of equal surf 04^.

1ÂŁ a sphere and a regular polyedron have the same center,
and equal surfaces ; each of the faces of the polyedron must
fidl partly wit/dn the sphere. For the solidity of a circum-
scribing solid is greater than the solidity of the sphere, as the
one includes the other : and therefore, by the preceding arti-
cle, the surface of the former is greater than that of the
latter.

But if. the faces of the polyedron fall partly within the
sphere, their perpendicular aistance from the center must be
iess than the radius. And therefore, if the surface of the
polyedron be gnly equal to that of the sphere, its solidity
must be less. For the soliditv of the polyedron is equal to
its surface multiplied into | of the distance from the center.
(Art. 59.) And the solidity of the sphere is equal to its surface
multiplied into ^ of the radius.

Cor. A sphere has a less surface than any regular poly-
edron of the same capacity.

For other cases of Isoperimetry, see Fluxions.

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72

APPENDIX.â€” PART I.

CONTAINING RULES, WrmOUT I>BM0N8TIUTI0NS, FOR THE BIENSURATION
OP THE CONIC SECTIONS, AND OTHER FIGURES NOT TREATED OF IN-THE
ELBMENTS 09 EUCUD.*

PROBLEM I.

To find the area of an ellipss.

101. Multiply the product of the transverse and conjugate
axes into .7864.

Ex. What is the area of an ellipse whose transverse axis
is 36 feet, and conjugate 28 ? Ans. 791.68 feet.

PROBLEM II.

To fiiid the area of a segment of an ellipse, cut off by a
line perpetidictdar to either axis.

102. If either axis of an ellipse be made the diameter of a
circle ; and if a line perpendicular to this axis cut off a seg-
ment from the ellipse, and from the circle ;

The diameter of the circle, is to the other axis of the ellipse ;
As the circular segment, to the elliptic segment.

' ' ' â–  ' " T

â€˘ For demonstrations of these rules, see Conic Sectioii% Spheiical Tngonom-
etry, and Fliudons^ or Hutton's Mensuration.

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APPENIÂ»IX. 73

Ex. What is the area of a segment cut <^from an ellqise
whose truisverse axis is 415 feet, and conjugate 332 ; if the
heifi;fat of the s^m^it is 96 feet, and its base is perpendicular
to me transverse axis?

The circular segment is 23680 feet
And the elUptic si^m^it 18944

PROBLEM III.

To find the area of a conic vakabola.

103. Multiply the base by | of the height.

Ex. If the base of a parabola is 26 inches, and the height
9 feet ; what is the area ? Ans. 13 feet

PROBLEM IV.

To find the area of a frustum of a parabola^ cutoff by a
line parallel to the base,

104. Divide the difference of the cubes of the diameters of
the two ends, by the difference of their squares; and multiply
the quotient by | of the perpendicular height

Ex. What is the area of a parabolic frustum, whose height
is 12 feet, and the diameters of its ends 20 and 12 feet ?

Ans. 196 feet

PROBLEM V.

To find the area of a conic hyperbola.

105. Multiply the base by f of the height ; and correct the
product by subtracting from it the series

* ( fr Â» the hase or double ordinate,
In which < AÂ«>the h^[ht or abseissa,

( a^â€” the height divided by the sumof diehaight
and trans^rse axis.

10

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1

Thd Mites c on v e rge s so nqpidly, thsi a few of fiie^ first
temis will feDerdlly pre the correcticm wUh sufficient exact*
iteas. This correction is the difference between the b3rpei^
boki and a pantbola of t|ie same base and hetgfat

Ex. If tibs bMNi ot a hyperbola be 24 feet, the height 10
andthetransreiseaxisSO; what is the area?

The base x | tfie height is

160.

The first term of the series is
Thesec(Mid
The third
The fourth

0.016669
0.000692
0.000049

aoooooe

Th^sun

0.017313

This mto 2bh ia

8.31

And the area corrected ia

151.69

PBOBLEM VI.

T^JM the area o/ a sfheeical triangle formed hy three
arce (^ great circles of a sphere^

106. As 8 right angles or 720^,

To <he excess of ito 3 given angles above 180^ ;
So IS the Di^la swÂŁios <tf the qpiiere^
To iSm 9rea of the spherical triangle.

Ex. What k the area of a q>herical triangle, on a sphere
whose diameter is 30 fe^ if ttÂ»^ loigles are 130<^, 102^, and
68^? Ans. 471.24 feet

FÂ«e9&BM Til.

Te find the area of a spherical pol yoon fermed by arce
ef great drdee.

107. As 8 right angles, or 720^,

To the %Mess of all the ^peeeik Â«Â«|ies abote the pro-
duct of the imiidbÂ«r ^an rieii^ . if gitQ IM^ ;
Cb te M^ wIk^ swrftce of tte ^have^
To die area^Â«iÂ» qrtMtoal fetyg^m.

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on a sfhete whose diaiotfter is 17 ii^hes ; if the gum oÂŁ all
the angles is lOBOPI Am. 227 inches.

^mawLwrn Ytiu

Thjind thehmaramrface included b e i m\$m iwegrc^t â‚¬Me\$
tfaep/un.

lOS. As 3\$(P| to tbe angle made by the giiren oirtles;
So is the whole suriace of the sph^e^ to ihe siur&oe
between the circles.

Or,

The lunar sur&ce is equal to the breadth of ttit ttdddte
part of it| multiplied into the diameter of the sphere.

Ex. If the earth be 7930 miles in diameter, what is die
surface of that part of it which is inchided between th0 SAOi
and 83d d^ee of longitude )

Ans. 9,878,000 s^piare ntited.

PROBLEM IX.

To fold the selidUtf of a epmwnDf/ormed by the revolution
of an ellipse about either axis.

109. Multiply the product of the fixed axis and the square
of the levolving axis, into .6236.

Ex. 1. What is the solidity of an obloqg n>heroid, whose
longest and shortest diameters are 40 and W teeil

Ans. 40x30Â»x.6236-18850feet

2. If the earth be an oblate spheroid, whose polar and
equatorial diameters are 7930 and 7960 miles ; what is its
solidity ? Aim. 963,000,000,000 miles.

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^ AFPSMBISL-

PROBLEM X.

To find the selidiif/ of the middle frustum of a spheroid^
included between two planes which are perpendicular to
the axisj and equally distant from-the center.

110. Add toge^thet the square of the diameter of <me end;
and twice the square of the middle diameter ; multiply the
sum by \ of the height, and the product by .7864.

If P and (2= the two diameters, and h^the height ,*
The solidity-(2DÂ»+rfÂ«)xiAx.7854.

Ex. If the diameter of one end of a middle frustum of a
qtercid be 8 inches, the middle diameter 10 and the height
30, what is the solidity ?

Ans. 2073.4 inches.

Coy. Half the middle frustum is equal to a frustum of
which one of the ends passes through the center.
If th^i D and c(->the diameters of the two ends^ and A=the

height,

The solidityÂ«(2DÂ«+dÂ«)XiAx.7854.

PROBLEM XI.

To find the solidity of a paraboloid.

111. Multiply the area of the base by half the height.

Ex. If the diameter of the base of a paraboloid be 12 feet,
and the height 22 feet, what is the solidity ?

Ans. 1243 feet.

problem XII.

To find the solidity of a frustum <jf a paraboloid.

112. Multiply the sum of the areas of the two ends by half
their distance.

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API>ENBIX.

77

Ex. If the diameter of one end of a frustum of a parabo-
loid be 8 feet, the diameter of the other end 6 feet, and the
length 24 feet ; what is the soUdity }

Ans. 942ifeet.

Cor. If a cask be in the form of two equal frustums of a
paraboloid ; and

If D=the middle diam. rf=the end diam. and A==the length ;
The solidity=(DÂ»+dÂ«)xiAx.7854.

PROBLEM XIII.

To find the solidity of a hyperboloid, produced by the
revolution of a hyperbola on its axis.

113. Add together the square of the radius of the base, and
the square of the diameter of a section which is equally dis-
tant from the base and the vertex ; multiply the sum by the
height, and the product by .5236.

If R=the radius of the base, D=the middle diameter, and
A=the height ;

The solidity=:(R^-f-DÂ«)xAx.6236.

Ex. If the diameter of the base of a h3rperboloid be 24, the
square of the middle diameter 252, and the height 10, what
is the solidity? Ans. 2073.4.

PROBLEM XIV.

To find the solidity of a frustum of a hyperboloid.

114. Add together the squares of the radii of the two ends,
and the square of the middle diameter ; multiply the sum by
the height, and the product by .5236.
If R and rÂ«=the two radii, D=the middle diameter, and

h^itie height;

The solidity=(RÂ«+rÂ«+D2)xAx.5236.

Ex. If the diameter of one end of a frustum of a hyperbo-
loid be 32, the diameter of the other end 24, the square of the
middle diameter 793\$, and tibie length 20, what is the solidity ?

Ans. 12499.3.

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78

PROBLEM XV.

To find the solidity of a circular spindlEi produced by
the revolution of a circular segment about its base or
chord as an axis.

116. Prom \ of the cube of half the axis, subtract the pro-
duct of the central distance into half the revolving circular
s^fment, and multiply the remainder by four times 3.14169.

If aÂ»the erea of the revolving circular segment,
Z=half the length or axis of the spindle,
cÂ»the distance of this axis from the cent^ of the
circle to which the revolving segment belongs ;

The solidity- (iZ3_jac)x4x3.14159.

Ex. Let a circular spindle be produced by the revolution
of the segment ABO (Fiff. 9.) about AB. If the axis AB be
140, and OP half the middle diameter of the spindle be 38.4;
what is the solidity?

The area of the revolving segment is 3791
The central distance PC 44.6

The solidity of the spindle 374402

PROBLEM XVI.

To find the solidity of the middle frustum of a circular

spindle,

116. Prom the square of half the axis of the whole spindle,
subtract | of the square of half the len|?th of the frustum ;
multiply the remainder by this half lÂ«[igth ; from the product

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