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For, since AP AV
By division AP
That is,

:: AV; BV;

AP-AV :: AV : AV-BV;


Hence, VP. AV = AP . AB.

But VP. AV is either

and th. 23 of this chapter).

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:: AV AB.

or < AP2 (cor. to th. 31 Geom.

Therefore AP ABAP2: whence 4ABAP, or AP4AB. Consequently, the minimum value of AP is the quadruple of AB; and in that case PV VA = 2AB.


Q. E. D*.

Of all Right Cones Circumscribed about the Same Sphere, the Smallest is that whose Altitude is Double the Diameter of the Sphere.


For, by th. 30, the solidity varies as (see the fig. to that theorem): and, by lemma 2, since VA - VB is given, the third proportional is a minimum when VA=2AB. Q.E.D.



Cor. 1. Hence, the distance from the centre of the sphere to the vertex of the least circumscribing cone, is triple the radius of the sphere.

Cor. 2. Hence also, the side of such cone is triple the radius of its base.


The Whole Surface of a Right Cone being Given, the Inscribed Sphere is the Greatest when the Slant Side of the Cone is Triple the Radius of its Base.

For, let c and c' be two right cones of equal whole surface, the radii of their respective inscribed spheres being

*Though the evidence of a single demonstration, conducted on sound mathematical principles, is really irresistible, and therefore needs no corroboration; yet it is frequently conducive as well to mental improvement, as to mental delight, to obtain like results from different processes. In this view it will be advantageous to the student, to confirm the truth of several of the propositions in this chapter by means of the fluxional analysis. Let the truth enunciated in the above Jemma be taken for an example: and let AB be denoted by a, av by x, вV by x-a. Then we shall have x-a:x:: x : the third proportional; which is to be a minimum. Hence, the fluxion of this fraction will be equal to zero (Flux. art. 51). That is (Flux. = o. Consequently, aa—2ax = 0, and x=2α,

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x2x-2axx (x-a)2. er AV = 2AB, as above.

arts. 19 and 30),


denoted by R and R'; let the side of the cone c be triple the radius of its base, the same ratio not obtaining in c; and let c' be a cone similar to c, and circumscribed about the same sphere with c'. Then, (by th. 31) surf. c'< sarf. c': therefore surf. c'‹ surf c. But c" and c are similar, therefore all the dimensions of c" are less than the corresponding dimensions of c: and consequently the radius R' of the sphere inscribed in c" or in c', is less than the radius R of the sphere inscribed in c, or R > R ́. Q. E. D.

Cor. The capacity of a right cone being given, the inscribed sphere is the greatest when the side of the cone is triple the radius of its base.

For the capacities of such cones vary as their surfaces (th. 29).


Of all Right Cones of Equal Whole Surface, the Greatest is that whose Side is Triple the Radius of its Base: and reciprocally, of all Right Cones of Equal Capacity, that whose Side is Triple the Radius of its Base has the Least Surface.

For, by th. 29, the capacity of a right cone is in the compound ratio of its whole surface and the radius of its inscribed sphere. Therefore, the whole surface being given, the capacity is proportional to the radius of the inscribed sphere: and consequently is a maximum when the radius of the inscribed sphere is such; that is, (th. 32) when the side of the cone is triple the radius of the base*.

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.

* Here again a similar result may easily be deduced from the method of fluxions. Let the radius of the base be denoted by 1, the slant side of the cone by z, its whole surface by a2, and 3.141593 by π. Then the circumference of the cone's base will be 27x, its area 1a, and the convex surface The whole surface is, therefore, a2 + пiz: and this being



· α,

we have z=—-X. But the altitude of the cone is equal to the square root

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of the difference of the squares of the side and of the radius of the base; that



is, it is W- -). And this multiplied into of the area of the base,

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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 II.


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

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P', surf. P < surf. p'.

Ist. If surf. P = surf. P', P > P'. 2dly. If . . P = 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).


this being a maximum its square must be so likewise (Flux. art. 53), that is, *r* 270214 -, or, rejecting the denominator, as constant, a1a— 2πa214 must be a maximum. This, in fluxions, is 2atxi—8πa2±3r = 0; whence we have a2 247120, and consequently r N

47; and a2 = 4π2. Substituting

this value of a for it, in the value of z above given, there results. - 1 = 41 - 13r. Therefore, the side of the cone

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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.


But, Ist. surf. P: surf. c:: surf. p': surf. c'; whence, if surf. P= surf. P', surf. c therefore c> c'. But p:c:: P': c'.


surf. c'; Therefore P > P'.

2dly. PC:: P: c'. Theref. if P=P', c=c': consequently surf. c < surf. c'. But, surf. p: surf. c:: surf. r' : surf. c'. Whence, surf. p < 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.


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 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. sequently, 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 circumscribing the same sphere, are respectively as their surfaces.

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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 spheres: 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 P 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 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 below.


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.

Er. 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 perimeter of an irregular polygon which circumscribes it is 15: what are their respective areas?

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