Again, reciprocally, the capacity being given, the surface is in the inverse ratio of the sphere inscribed therefore, it is the smallest when that radius is the greatest; that is (th. 32) when the side of the cone is triple the radius of its base. Q. E. D.

THEOREM XXXIV.

The Surfaces, whether Total or Lateral, of Pyramids Circumscribed about the Same Right Cone, are respectively as their Solidities. And, in particular, the Surface of a Pyramid Circumscribed about a Cone, is to the Surface of that Cone, as the Solidity of the Pyramid is to the Solidity of the Cone; and these Ratios are Equal to those of the Surfaces or the Perimeters of the Bases.

For, the capacities of the several solids are respectively as their bases; and their surfaces are as the perimeters of those bases so that the proposition may manifestly be demonstrated by a chain of reasoning exactly like that adopted in theorem 11.

04

cone is equal to the square root of the difference of the squares of the side and of the radius of the base; that is, it is =

203

T

And this multiplied into of the area of the base, viz. by **,gives

2a2

―), for the capacity of the cone. Now, this being

a maximum its square must be so likewise (Flux. art. 53), that is,

9

-

, or, rejecting the denominator, as constant, aaxa 2xα2x4 must be a maximum. This, in fluxions, is 2a xx-8=a3x3 *

a2

4

o; whence we have a2-4x2 = o, and consequently x = ✔~; and @2 = 4x3. Substituting this value of a2 for it, in the value of =

above given, there results z=

47x2

-x=4x-x = 31.

Therefore, the side of the cone is triple the radius of its base. Or, the square of the altitude is to the square of the radius of the base, as 8 to 1, or, to the square of the diameter of the base, as 2 to 1.

THEOREM

THEOREM XXXV.

The Base of a Right Pyramid being Given in Species, the Capacity of that Pyramid is a Maximum with the Same Surface, and on the contrary, the Surface is a Minimum with the Same Capacity, when the Height of One Face is Triple the Radius of the Circle Inscribed in the Base.

LET P and P be two right pyramids with similar bases, the height of one lateral face of P being triple the radius of the circle inscribed in the base, but this proportion not obtaining with regard to P': then

1st. If surf. P = surf. p', p > P'.

2dly, If. P= P', surf. P < surf. p'.

surf. Þ': surf. c';

For, let c and c' be right cones inscribed within the pyramids P and P' then in the cone c, the slant side is triple the radius of its base, while this is not the case with respect to the cone c'. Therefore, if c = = c, surf. c < surf. c and if surf. c = surf. c', c> c' (th. 33). But, 1st. surf surf c whence, if surf. p = surf. p' surf. c = surf. c'; therefore c> c. 2dly, PC:: P' surf. c< surf. c'. Whence, surf. P<

But P c :: P: c'. Therefore P > P'.
c'. Theref. if PP', c=c': consequently
But, surf, P surf. c:: surf. P' surf. c.
surf. P'

Cor. The regular tetraedron possesses the property of the minimum surface with the same capacity, and of the maximum capacity with the same surface, relatively to all right pyramids with equilateral triangular bases, and, a fortiori, relatively to every other triangular pyramid.

THEOREM XXXVI.

A Sphere is to any Circumscribing Solid, Bounded by Plane Surfaces, as the Surface of the Sphere to that of the Circumscribing Solid.

For, since all the planes touch the sphere, the radius drawn to each point of contact will be perpendicular to each respective plane. So that, if planes be drawn through the centre of the sphere and through all the edges of the body, the body will be divided into pyramids whose bases are the respective planes, and their common altitude the radius of the sphere. Hence, the sum of all these pyramids, or the whole circumscribing solid, is equal to a pyramid or a cone whose

base,

base is equal to the whole surface of that solid, and altitude equal to the radius of the sphere. But the capacity of the sphere is equal to that of a cone whose base is equal to the surface of the sphere, and altitude equal to its radius. Consequently, the capacity of the sphere, is to that of the circumscribing solid, as the surface of the former to the surface of the latter both having in this mode of considering them, a common altitude.

:

Q E. D.

Cor. 1. All circumscribing cylinders, cones, &c. are to the sphere they circumscribe, as their respective surfaces. For the same proportion will subsist between their indefinitely small corresponding segments, and therefore between their wholes.

Cor. 2. All bodies circamscribing the same sphere, are respectively as their surfaces.

THEOREM XXXVII.

The Sphere is Greater than any Polyedron of Equal Surface.

For, first it may be demonstrated, by a process similar to that adopted in theorem 9, that a regular polyedron has a greater capacity than any other polyedron of equal surface. Let P, therefore, be a regular polyedron of equal surface to a sphere s. Then P must either circumscribe s, or fall partly within it and partly out of it, or fall entirely within it. The first of these suppositions is contrary to the hypothesis of the proposition, because in that case the surface of P could not be equal to that of s. Either the 2d or 3d supposition therefore must obtain; and then each plane of the surface of P must fall either partly or wholly within the sphere s : whichever of these be the case, the perpendiculars demitted from the centre of s upon the planes, will be each less than the radius of that sphere and consequently the polyedron must be less than the sphere s, because it has an equal base, but a less altitude.

:

Q. E. D.

Cor. If a prism, a cylinder, a pyramid, or a cone, be equal to a sphere either in capacity, or in surface; in the first case, the surface of the sphere is less than the surface of any of those solids; in the second, the capacity of the sphere is greater than that of either of those solids.

The theorems in this chapter will suggest a variety of practical examples to exercise the student in computation. A few such are given in the following page.

EXERCISES

EXERCISES.

Ex. 1. Find the areas of an equilateral triangle, a square, a hexagon, a dodecagon, and a circle, the perimeter of each being 36.

Ex. 2. Find the difference between the area of a triangle whose sides are 3, 4, and 5, and of an equilateral triangle of equal perimeter.

Ex. 3. What is the area of the greatest triangle which can be constituted with two given sides 8 and 11: and what will be the length of its third side?

Ex. 4. The circumference of a circle is 12, and the pèrimeter of an irregular polygon which circumscribes it is 15: what are their respective areas?

Ex. 5. Required the surface and the solidity of the greatest parallelopiped, whose length, breadth, and depth, together make 18?

Ex. 6. The surface of a square prism is 546: what is its solidity when a maximum ?

Ex. 7. The content of a cylinder is 169-645968: what is its surface when a minimum ?

Ex. 8. The whole surface of a right cone is 201·061952: what is its solidity when a maximum ?

Ex. 9. The surface of a triangular pyramid is 43.30127: what is its capacity when a maximum ?

Ex. 10. The radius of a sphere is 10. Required the solidities of this sphere, of its circumscribed equilateral cone, and of its circumscribed cylinder.

Ex. 11. The surface of a sphere is 28-274337, and of an irregular polyedron circumscribed about it 35: what are their respective solidities?

Ex. 12. The solidity of a sphere, equilateral cone, and Archimedean cylinder, are each 500: what are the surfaces and respective dimensions of each?

Ex. 13. If the surface of a sphere be represented by the number 4, the circumscribed cylinder's convex surface and whole surface will be 4 and 6, and the circumscribed equilateral cone's convex and whole surface, 6 and 9 respectively. Show how these numbers are deduced.

Ex. 14. The solidity of a sphere, circumscribed cylinder, and circumscribed equilateral cone, are as the numbers 4, 6, and 9. Required the proof.

PROBLEMS

PROBLEMS RELATIVE TO THE DIVISION OF FIELDS OR OTHER SURFACES.

PROBLEM I

To Divide a Triangle into two parts having a Given Ratio,

m: n.

1st. By a line drawn from one angle

of the triangle.

Make AD AB : : m : m + n; draw CD.
So shall ADC, BDC, be the parts required.

Here, evidently, AD =

m

m+n

AB, DB

m+n

AB.

2dly. By a line parallel to one of the sides of the triangle. Let ABC be the given triangle, to be divided into two parts, in the ratio of m to n, by a line parallel to the base AB. Make CE to EB as m to n: erect ED perpendicularly to CB, till it meet the semicircle described on cв, as a diameter, in Ꭰ. Make CF CD and draw through F, GF | Ab. divide the triangle ABC in the given ratio.

CD2

A

So shall GF

For, CE: CB=—-: :: CD2 (= CF2): CB2. But CE : EB :: m :n,

CE

or CE : CB: mm+n, by the construction: therefore CF2: CB2 :: m : m+n. And since ▲ CGF : ▲ CAB :: CF2: CB2 ; it follows that CGF: CAB :: m : m+n, as required. Computation. Since CB2: CF3 :: m + n : m, therefore, (m + n) cr2 = m . CB2; whence CF ✔ (m + n)

CF CB

m

In like manner, CG = CA √

m+n 3dly. By a line parallel to a given line. Let ni be the line parallel to which a line is to be drawn, so as to divide the triangle ABC in the ratio of m

to n.

D

= CB √m, or

m

m+n

By case 2d draw GF parallel to AB, so as to divide ABC in the given ratio. Through F draw FE parallel to HI. On CE as a diameter describe a semicircle; draw GD perp. to AC, to cut the semicircle in D. Make CP CD : through P, parallel to EF, draw PQ, the line required.

B

The