Jeremiah Day.

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Font size of a right cone, (Art. 65.) and UCY the sur&ce of a portioa
of the cone, cut off by a plane paralld to the base. Then
will ABCD be the surface (^the firustam.

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48 MENSURATION OF TmS CONE.

Then the area ABV=iax(A+d)â€” ioA+iorf. (Art. 34.)

And the area DCV=i6d.

Subtracting the one from the other,

But d : d'\-h ::b:a. (Sup. Euc. 8. 1.) Therefore iociâ€” J6*-i6A,

The surface of the frustum, then, is equal to

iah+ibh. or ihx{a+b)

Ck)r. The surface df the frustum is equal to the product of
the slant-height into the circumference of a circle which is
equally distant from the two ends. Thus, the surface ABCD
(Fig. 23.) is equal to the product of AD into MN. For.MN
IS equal to half the sum of AB and DC.

Ex. 1. What is the Convex -surface of a frustum of a rig^
cone, if the diameters of the two ends be 44 and 33, and the
slant-height 84? Ans. 10169.8.

2. If the perpendicular height of a frustum of a right cone
be 24, and the diameters of the two ends 80 and 44, vfhpX is
the whole surface ?

Half the difference of the diameters is 18v

And "^ 18Â»+24*=30, the slant-height, ( Art. 52.)
The convex surface of the frustum is 5843

The sum of the areas of the two ends is 6547

And the whole sur&ce is 12390

PROBLEM VI.

To find the solidity of a frustum of a cone.

68. Add together the areas of the two ends, and
the sauare root of the product of thesfi areas; and
multiply the sum by j of the perpendicular height.

This rule, which was given for the frustum of a pyramirf,
(Art. 50.) is equally applicable to the frustum of a cone \ be-
cause a cone and a pyramid which have equal bases and
altitudes are equal to each other.

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MEN8UBATION OF THE SPHBRR 49

Ex. 1. What is the solidity of a mast which is 72 feet longj
2 feet in diameter at one end, and 18 inches at the other ?

Ans. 174.36 cubic feet

2. What is the capacity of a conical cistern which is 9 feet
deep, 4 feet in diameter at the bottom, and 3 feet at the top ?

Ans. 87.f8 cubic feet=652.15 wine gallons.

3. How many gallons of ale can be put into a vat in the
form of a conic mistum, if the larger diameter be 7 feet, the
smaller diameter 6 feet, and the depth 8 feet ?

PROBLEM VIIÂ«

To find the surface of a spheiie.

69. MtJLflPLY THE diameter BV THE CIRCUMFERENCE.

Let a hemisphere be described by the quadrant CPD^
(Fig. 25.) revolving on the line CD. Let AB be a side of a
regular polygon inscribed in the circle of which DBP is an
arc. Draw AO and BN perpendicular to CD, and BH per-
pendicular to AO. Extend AB till it meets CD continued.
The triangle AOV, revolving on OV as an axis, will describe
a right cone. (Defin. 2.) AB will be the slant-height of a
frtLstum of this cone extending from AO to BN. Prom Q
the middle of AB, draw GM parallel to AO. The surface of
the frustum described by AB, (Art. 67. Cor.) is equal to

ABxctrcOM.*

Prom the center C draw CG, which will be peipendicular
to AB, (Euc. 3. 3.) and the radius of a circle inscribed in the
polygon. The triangles ABH and CGM are similar, because
the sides are perpencUcular, each to each. Therefore,

HB or ON : AB : : GM : GO : : circ GM : circ GO.

So that ONxcirc GC=*ABxcirc GM, that is, the surface
of the frustum is equal to the product of ON the perpendicu-
lar height, into cire GO, the perpendicular distance from the
center of the polygon to one of the sides.

7

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go MENSURATION OF THE SPHERS.

In the same manner it may be proved, that the sur&ces
produced by the revolution of the lines BD and AP about the
axis DC, are equal to

ND xcirc GC, and 00 xcirc GO.

The surfiu^ of the whole solid, therefore, (Euc. 1.2.) is equal to

OD xcirc GO.

The demonstration is applicable to a solid produced by
the revolution of a polygon of anj/ number of sides. But a
{wlygon may be supposed which shall differ less than by any
fiiven quantity from the circle in which it is inscribed; (Sup.
ÂŁuc. 4. 1.) and in which the perpendicular GO shall differ .
less than by any given quantity from the radius of the circle.
Therefore, the surface of a hemisphere is equal to the product
of its radius into the circumference of its base ; and the sur-
face of a sphere is equal to the product of its diameter into
its circumference.

Oor. 1. From this demonstration it follows, that the surface ^*
of any segment or zone of a sphere is equal to the product of "'
the height of the segment or zone into the circumierence of
the sptere. The surface of the zone produced by the revolu-
tion of the arc AB about ON, is equal to ON xcirc OP. And
the surface of the segment produced by the revolution of BD
about DN is equal to DNxcirc OP.

Oor. 2. The surface of a sphere is equal to four times the
area of a circle of the same diameter ; and therefore, the con-
vex surface of a hemisphere is equal to twice the area of its
base. For the area of a circle is equal to the product of half
the diameter into half the circumference ; ^Art 3W.) that is,
to ^ the pr6duct of the diameter and circumterence.

Oor. 3. The surface of a sphere, or the convex surfiM^ of
any spherical segment or zone, is equal to that of the circum-
tcnbing C3^inder. A hemisphere described by the revdution
of the arc DBP, is circumscribed by a cyliiKier produced by
the revolution of the paraMogran^ DrfOP. The convex but-
&ce of the cylinder is equal to its height multiplied bv its
cireumfOTOTce. (Art. 62.) And this is also the surfece ot the
heDoisphere*

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MBNflinATION OP THE ,8P^SIUS. 5|

So ibe surface produced by the revolution of AB is equal
to that produced by the revolution of ab. And the surÂŁatce
produced by BD is equal to that produced by b(L

Ex. 1. Considering the earth as a sphere 7930 miles in di-
ameter, how many square miles are there on its surface ?

Ans. 197,558,500.

2. If the cir<^mference of the sun be 2,800,000 miles, what
is his surface ? Ans. 2,495,547,600,000 sq. miles*

. 3. How many square feet of lead will it require, to cover a
-hemispherical oome whose base is 13 feet across ?

Ans. 265^.

PROBLEM VIII.

To find the solidity of a sphere.

70. 1. Multiply the cube op the diameter by â€˘5236.
â€˘ â€˘r Or,

2. Multiply the square op the diameter by | op
the circumference.

Or,

3. Multiply the surface by | op toe diameter.

1. A sphere is two thirds of its circumscribinff cylinder.
(Sup. Euc. 21. 3.) The height and diameter of the cylinder
are each equal to the diameter of the sphere. The solidity
of the cylinder is equal to its height multiplied into the area
of its base, (Art. 64.) that is putting D for the diameter,

DxDÂ«x.7854 or I>^xJ85i.

And the solidity of the sphere, being | of this, is

D'X.5236.

2. The base of the circumscribing cylinder is equal to half
the-circumference multiplied into hdf the diameter ; (Art. 30.)
that is, if C be put for the circumference,

iCxD; and the solidity is jCxD*.

Therefore, the solidity of the sphere is

fof iCxD'-D^xiC.

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02 MBNSDRATION OF THB SPllBRE.

S. hi the last expression, which is the same as CxDx^D,
we may substitute S, the surface, for CxD. (Art 69.) We
then have the solidity of the sphere equal to

Sx^D.

Or, the sphere may be supposed to be filled with small
pyramids^ standing on the surface of the sphere, and having
their common vertex in the center. The number of these
may be such, that the difference between their sum and the
sphere shall be less than any given quantity. The solidity
of each pyramid is equal to the product of its base into i of
its height. (Art. 48.) The solidity of the whole, therefore, is
equal to the product of the surface of the sphere into | of its
raidius, or | of its diameter.

71. The numbers 3.14169, .7854, .6236, should be made
perfectly familiar. The first expresses the ratio of the cir^
cumference of a circle to the diameter; (Art. 23.) the second,
the ratio of the area of a circle to the square of the diameter
(Art. 30.)", and the third, the ratio of the solidity of a sphere
to the ctibe of the diameter. The second is i of the first, and
the third is | of the first.

As these numbers are frequently occurring in mathematical
investigations, it is common to represent the first of them by
the Greek letter Â«â€˘. According to this notation,

'-3.14159, i' - 7854, |'-.5236.

If D"Â» the diameter J and R^^ the radius of any circle or sphere ;

Then, D=2R D2=^4RÂ» D3^8R^

And ir D ; .. ^.r J Â«â€˘ DÂ« ; - the area of | Â«-D Â» ; ,.
Or, 2^R \ =^^^ ^^*^*orxRÂ« \ the circ. or J^R' j = *Â«
solidity of the sphere,

Ex. 1. What is the solidity of tfie earth, if it be a sphere
7930 miles in diameter ?

Ans. 861,107,000,000 cubic miles.

2. How many wine gallons will fill a hollow sphere 4 feet
in diameter ?

Ans. The capacity is 33.5104 feet-=260| gallons.

3. If the diameter of the moon be 2180 miles, what is its
â€˘olidity ? Ans. 6,424,600,000 mileg.

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MINSUlATION OFTHB BPIHRB. 53

72. If the solidity of a sphere be given, the diameter may
be found by reversing the first rule in the preceding article ;
that is, dividing by .6236 and extracting the cube root of the
quotient

Ex. 1. What is the diameter of a sphere whose solidity is
66.46- cubic feet? Ans. 6 feet.

2. What must be the diioneter of a globe to contain 16766
pounds of water ? Ans. 8 feet.

PROBLEM IX.

To find the convex surface of a segment or zone of a

sphere.

73. Multiply the height op the segment or zone

INTO the circumference OP THE SPHERE.

For the demonstration of this rule, see art. 69.

Ex. 1. If the earth be considered a perfect sphere 7930
miles in diameter, and if the polar circle be 23Â° 28' from the
pole, how many square miles are there in one of the frigid
zones ?

If PQOE (Fig. 16.) be a meridian on the earth, ADB one
of the polar curcles, and P the pole ; then the frigid zone is a
spherical segment described by the revolution of the arc APB
about PD. The angle AOD subtended by the arc AP is 23*^
28', And m the right angled triangle ACD,

R : AC : : cos ACD : CD^363r.

Then, CP~CDÂ«=3966-^637^338Â«PD the height of
the segment.

And 328x7930x3.14169=8171400 the surface.

2. If the diameter of the earth be 7930 miles, what is the
surface of the torrid zone, extending 23^ 28' on each side of
the equator ?

If Ed (Fig. 16^ be the equator, and GH one of the tropics,
then the angle EGG is 23^ 28'. And in the right angled
triangle GCM,

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fl lODmnuncmoF

R : OG : : sin ECG : GMÂ»Ca^Â«-1678.9 the beighl of
hatf the zone.

The surface of the whole zone is 78669700.

3. What is the surfiM)eof each of the taoiperate 2ones?
The height DNÂ«CPâ€”CNâ€”PD= 2058.1

And the surface of the zone is 51273000.

The surface of the two temperate zones is 102,646,000
of the two fri^ zones 16,342,800

of the torrid zone 78,669,700

of the whole globe 197,558,500

PROBLEM X.

To find the solidity of a spherical sector.
74. Multiply the spherical surface by | of ths

The spherical sector, (Fig. 24.) produced by the revolution
of ACBD about CD, may be supposed to be filled with smaU
pyramids^ standing on the spherical surface ADB, and ter-
minating in the point C. Their number may be so great,
that the height of each shall difler less Ihan by any given
length from the radius CD, and the sum of their bases shall
dii^r less than by any given quantity from the surface ABD.
The solidity of each is equal to the product of its base into I
of the radius CD. (Art. 48.) Therefore, the solidity of all ot
them, that is, of the sector ADBC, is equal to the pieduet of
the spherical surface into | of the radius.

Ex. Suppofi^ the earth to be a sphere 7930 miles in di-
ameter, and the polar circle ADB (Fig. 16.) to be 23^ 28^
from the pole ; what is the solidity of the spherical sector
ACBP? Ans. 10,799,867,000 miles.

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iBHSUllATlON OF THB SPREBSl 55

PROBLEM XI.

To find the solidity of a spherical segment.
75. Multiply half the height op the segment

INTO the area op THE BASE, AND THE CUBE OP THE
HEIGHT INTO .52^36 ; AND ADD THE TWO PRODUCTS.

As the circular sector AOBC (Fig. 9.) consists of two parts,
the segment AOBP and the trian^e ABC ; (Art. 35.) so the
spherical sector produced by the revolution of AOC about
OC coixsists of two parts, the segment produced by the revo-
lution of AOP, and the cone produced by the revolution of
ACP. If then the cone be subtracted from the sector, the
remainder will be the segment.

Let CO=R, the radius of the sphere,

PB=r, the radius of the base of the segment,
POÂ«A, the height of the segment,
Then PC Â«=Râ€” rA, the axis of the cone. .

The sector=2^RxAxjR(Arts. 71,73, 74.)=|Â«-ARÂ«.
The coneÂ«!rr2xi(Râ€” A)(Arts. 71, 66.)=J?rr2R_iirArÂ».

Subtracting the one from the other,

The segment Â«*=|9r*Ra â€” >.srrÂ«R+jirArÂ».

But DOxPO==BOÂ« (Trig. 97.*)=PO"Â«+PB^ (Euc. 47. 1.)

That is, 2 RAÂ«. AÂ» +r^ . So that, RÂ«= ^JT'

4A*

Substituting then, for R and RÂ«, their values^ and nmltt-
plying the factors.

The s^mentÂ«|xA3+|irArÂ«+i^^^^ \^hr* â€” r'V+i*^*^'

"^xdiicti, by uniting the terms, becomes

* Euclid 31| 3, and 8, 6. Cor.

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gg MENSURATION GP THE fiPHIRS.

The first term here is 4AxÂ«-rS half the height of the s^-
ment multiplied into the area of the base ; (Art. 71.) and the
other A'xi*-, the cube of the height multiphed into .6236.

If the segment be greater than a hemisphere, as ABD ;
(Fig. 9.) the cone ABC must be added to the sector ACBD.

Let PD=A the heiffht of the segment.
Then, PC=^A â€” ^R me axis of the cone.

The sector ACBD - f Â«r ARÂ«

The etme=-Â«rrÂ«Xi(Aâ€” R)-i5rArÂ«â€” |irrÂ«R

Adding them together, we have as before,

The segmentâ€” fxARÂ«â€”JxrÂ«R+|^irArÂ«.

Cor. The solidity of a spherical s^ment is equal to half
a cylinder of the same base and height -f a sphere whose diÂ«
ameter is the hei^t of the segment. For a cylinder is equal
to its hdght multiplied into the area of its base ; and a sphere
is equal to the cube of its diameter multiplied by .5236.

Thus, if Oy (Fig. 16.) be half Oir, the spherical s^fment
produced by the revolution, of Oxi is equal to the cylinder
produced by ivyx + the sphere produced by Oyxz ; suppos-
mg each to revolve on the Ime 0:r.

Ex. 1. If the height of a spherical segment be 8 feet, and
the diameter of its l^ise 25 feet ; what is the solidity?

Ans. (25)Â»X.7854x4+8Â»x.5236-2231.58feet.

2. If the earth be a sphere 7930 miles in diameter, and the
polar circle 23Â° 28' from the pole, what is the solidity of one
of the frigid zones? Ans. 1,303,000,000 miles.

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MENSURMION OF THE SPHERE. 57

PROBLEM XII.

To find the soliditt of a spherical zonb or frustum.

76. From the solidity op the whole sphere, sub-
tract THE TWO segments ON THE SIDES OP THE ZONE.

Or,

TWO ENDS, AND | THE SQUARE OP THEIR DISTANCE ; AND
MULTIPLY THE SUM BY THREE TIMES THIS DISTANCE, AND
THE PRODUCT BY .5236.

If firom the whole sphere, (Fig. 15.) there be taken the two
s^ments ABP and GHO, there will remain the zone or frus-
tum ABGH.

Or, the zone ABGH is equal to the difference between the
segments GHP and ABP.

X)P=A ( *^Â® heights of the two segments.

;^^ / ^^^ radii of their bases.
AU=r )

DN=rf=H â€” A the distance of the two bases, or
the height of the zone.

Then the larger segment=>=iÂ«^HRÂ« + ^5rH3 Kt^ ^r x
And the smafler segment-i^rAr^+i^rAs ^ K^^- '^â€˘;

Therefore the zone ABGHÂ«|x(3HRa+H3â€” 3ArÂ»â€” A^)

By the properties of the circle, (Buc. 35> 3.)

ONxH=RÂ«. Therefore, (0N+H)xH=R2+HÂ«

RÂ«+H>

Or, 0P=

H

r^+A^

In the same manner, OP=â€” â€” ^ â€”

Therefore, 3Hx(rÂ«+AÂ«)=3Ax(R>+HÂ«.)
Or, 3HrÂ«+3HA* -3ARÂ«â€” 3AHÂ»-0. (Alg. 178.)

8

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Hg joawutunoN op thb

To refluce/he expression for the solidity of the zone to the
required form, without altering its value, let these terms be
added to it : and it will become

JÂ»(3HRÂ«+3HrÂ«â€” 3ARÂ«â€” 3ArÂ«+HÂ»â€” 3HÂ«A+3HAÂ« - AÂ»)

Which is equal to

j^x3(H - A)x(RÂ»+rÂ«+i(Hâ€” A)*)

Or, as !Â«- equals .6236 (Art 71.) and H â€” h equals c{,

The zone-^236x3rfx(RÂ«+rÂ»+Jrfa)

Ex. 1. If the diameter of one end of a spherical zcme is 24
feet, the diameter of the other end 20 feet, and th^ distance
of the two ends, or the height of the tone 4 feet ; what is die
.8(didity7 Ans. 1566.6 feet

2. If the earth be a sphere 7930 miles in diameter, and the
obliquity of the ecliptic 23Â° 28' ; what is the soUdity of one
of the temperate zones? Ans. 55,390,500,000 miles.

3. "What is the solidity of the torrid zone?

Ans. 147,720,000,000 miles.

The soUdity of the two temperate zcmes is 110,781,000,000
of the two frigid zones 2,606,000,000

of the torrid zone 147,720,000,000

of the whole globe 261,107,000,000

4. What is the convex surface of a spherical zone, whose
breadth is 4 feet, on a sphere of 25 feet oiameter?

6. What is the solHity of a spherical s^^ment, whose
height is 18 fe^ and the diameter of its base 40 feet?

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â– BMSimATIOll OP aOUDft

PROMISCVOUS EXAMPLES OF SOLIIXSU

09

Ex. 1. How much water can be put into a cubical vessel
three feet deep, which has been previously filled with cannon
balls of the same size, 2, 4, 6, or 9 inches in diameter, rega -
kurly arranged in tiers, one directly above another ?

Ans. 96 j wine gallons.

2. If a cone or pyramid, whose height is three feet^ be di-
vided into three equal portions, by sections parallel to the
base ; what will be the heights of the^ several parts ?

Ans. 24.961, 6.488, and 4.551 inches.

3. What is the solidity of the greatest square prism which
can be cut firom a cylindrical stick of timber, 2 fiaet 6 inches
in diameter and 56 feet long? *

Ans. 176 cubic feet.

4. How many such globes as the earth are equal in bulk
to the sun ; if the former is 7930 miles in diameter, and the
latter 890,000?

Ans. 1,413,678.

5. How many cubic feet of wall are fliere in a conical
tower 66 feet high, if the diameter of the base be 20 feet from
outside to outside, and the diameter of the top 8 feet ; the
thickness of the wall being 4 feet at the bottom, and decreas-
ing regularly, so as to be only 2 feet at the top ?

Ans. 7188.

â€˘ The common nile for measuring round timber w to multiply the square of
the quarter-girt by the length. The quarter-girt is one fourth of the croumfer-
ence. This method does not give the whole solidity. It makes an allowance of
about one-fifth, for waste in hewing, bark, &c The solidity of a cylinder is
equal to the product of the length into the area of the base.

If C = the circumference, and Â« = 3.14159, then (Art. 31.)
TheÂ«.Â«fthebMe = ^=(^)L (^-g^)'

If then the circumference were divided by 3.545^ instead of 4, and the quotient
squared, the area of the base would be correctly found. See note G.

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gQ MUMHURATION OF SOLIDS.

6. If a metallic globe filled with wine, which cost as much
at 5 dollars a gjallon, as the globe itself at 20 cents for every
square inch of its surface 5 what is the diameter of the globe?

Ans. 55.44 inches.

7. If the circumference of the earth be 25^000 miles, what
must be the diameter of a metallic globe, which, when drawn
into a wire ^V of an inch in diameter, would reach round the
earth ? Ans. 15 feet and 1 inch.

8. If a conical cistern be 3 feet deep, 7i feet in diameter at
the bottom, and 5 feet at the top ; what will be the depth of
a fluid occupying half its capacity ?

Ans. 14.535 inches.

9. If a ^lobe 20 inches in diameter be perforated by a cyl-
inder 16 inches in diameter, the axis of the latter passing
through the center of the jfcrmer ; what part of the solidity,
and the surface of the globe Will be cut away by the cylinder?

Ans. 3284 inches of the solidity, and 502,655 of the surface.

10. What is the solidity of the greatest cube which can be-
cut from a sphere three feet in diameter?

Ans. 51 feet.

11. What is the solidity of a conic frustum, the altitude of
which is 36 feet, the greater diameter 16, and the lesser di*
ameter 8 ?

12. What is the solidity of a spherical segment 4 feet high,
cut from a sphere 16 feet in diameter ?

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61

SECTION V.

I60PERIMETRY.*

â€˘ Art. 77. It is often necessary to compare a number of
different figures or solids, for tne pHirpose of ascertaining
which has the greatest area, within a given perimeter, or
the greatest capacity under a given surface. We may have
occasion to determine, for instance, what must be the form
of a fort, to contain a given number of troops, with the least
extent of wall ; or what the shape of a metallic pipe to con-
vey a given portion of water, or of a cistern to hold a given
quantity of liquor, with the least expense of msuterials.

78. Figures which have equal perimeters are called ho-
perimeters. When a quantity is greater than any other of
the same class, it is called a maximum, A multitude of
straight lines, of different lengths, may be drawn within a
circle. But among them all, the diameter is a marimtim.
Of all sines of angles, which can be drawn in a circle, the
sine of 90^ is a maarimum.

When a quantity is less than any other of the same class,
it is called a 7ninimum. Thus, of all straight lines drawn
from a given point to a ffiven straight line, tmit which is per-
pendicular to the given line is a minimum,. Of all straight
lines drawn from a given point in a circle, to the circumrar-
ence, the maximum, and minim^um, are the two parts of the
diameter which pass through that point. (Euc. 7, 3.)

In isoperimetry, the object is to determine, on the one
hand, in what cases the area is a maximum, within a given
perimeter ; or the capacity a maxim,um^, within a given sur-
lace : and on the other hand, in what cases the peiimeter is
a minimum for a given area, or the surface a minimum, for
a given capacity.

â€˘ Emerson*!} SiiDpson*tÂ» and Legendre's Geometry, Lhofllier, FonteneD^ Hnt-
ton't Mathmnaticii and Lond. PhiL Trans. Vol. 75.

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AQ laOPElllXETRV^

PROPOSITION I.

79. An Isosceles Triangle has a greater area than
any scalene triangle^ of equal base and perimeter.

If ABC (Fig. 26.) be an isoscdes trianffle whose equal sides
are AC and BC ; and if ABC be a scalene triangle on the
same base AB, and having AC'+BC'=AC+BC ; then the
area of ABC is greater than that of ABC.

Let perpendiculars be raised from each end of the base,
extend AC to D, make CD' equal to AC, join BD> and draw
CH and CW parallel to AB.

As the angle CAB=ABC, (Euc. 6, 1.) and ABD is a right
angle, ABC+CBDÂ«=CAB+CDB-ABC+CDB. Therefore
CHD=Â«CDB, so that CDâ€”CB; and by construction, CD'=-
AC. The perpendiculars of the equal right angled triangles
CHD and CHB are equal ; therefore, BH=}BD. In the
+BC=.D'C+BC. But D'C+BC>BD'. (Euc. 20, 1.)
^Aiy. But }BD, or BH, is the height of the isosceles trian-
gle; (Art.l.) and J AD' or AH', the height of the scalene
trian^e ; and the areas of two triangles which have the same
Jbase are as their heights. (Art. 8.) Therefore the area of
ABO is greater than that of ABC'. Among all triangles,
jtfaen, of a given perimeter, and upon a given Ixise, the isosce-
les triangle is a maximum.

Cor. The isoseeles triangle'has a less perimeter than any
ficalene triangle of the same base and area. The trianrie
ABC being less than ABC^ it is evident the perimeter of the
former must be enlarged, to make its. area equal to the area
of the latter.

PROPOSITION II.

80. A triangle in which two given sides maJce a right
ANGLE, has a greater area than any triangle in which the
same sides make an oblique angle.

If BC, BC, and BC" (Fig. 27.) be equal, and if BC be
perpeo^icular to AB ; thm ÂŁe right angled triangle ABC|

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ISOPBRllfBTBT. go

has a greater area than the acute angled triangle ABC, or
the oblique angled triangle ABC".

Let P'C am PC" be perpendicular to AP. Then, as the
three triangles have the same base AB, their areas are as their
heights ; that is, as the perpendiculars BC, PC, and PC*'.
But BC is equal to BC', and therefcMre greater than PC.
(Euc. 47. 1.) BC is also equal 4o BC", and tbeiehxe greats

PROPOSITION III.

81. IfaU the sides except one 4sf a polygon ke gUfen^

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