Chap. VIII.

GEOMETRY.

889

-lb 3 x 7 =14 x-— =14 x(a 3 + ub + b 3 ),

3 a — b 3 a-b 3 v p

Fig.1218.

Pyramid , or Cones: multiply the area of their base by onethird of their height for the solid content.

To find the solidity of the frustum of a cone or pyramid,add into one sum the area of the two ends, and the mean pro-portional between them, and take two-thirds of the sum fora mean area, which being multiplied by the perpendicularheight will give its content.

Suppose we call a 3 the area of the base of the frustum of apyramid, b 3 the area of the top, h the perpendicular height,and c the height of the pyramid when the frustum is madecomplete.

Then c + 4 = the height of the whole pyramid.

Then l a 3 (c + 4) is the content of the whole pyramid, and^ b 3 c the content of the top part: therefore the difference^ a 3 ( c + 4 ) — j 6* c is the content of the frustum.

But the quantity c being no dimension of the frustum, itmust be expelled from this formula, by substituting its valuein the following manner,

a 3 : b 3 :: (c+4) 8 : c 9 , or a : b :: c + 4 : c;lienee, a — b : b :: 4 : c, and a — b : a :: 4: c + 4;

. . /. bh . y ah

hence therefore, c=-r> an d c + 4 =-, ;

a — b a—b

then these values of c and c + 4 being substituted for them inthe expression for the content of the frustum gives that con-, „ ah ... bh a*-b 3

tent = J or x-;

J a — b

which is the rule above given, ab being the mean between a 3and b 3 .

Tetraedron. — Multiply the area of the base by onc-thirdof the perpendicular height for its solidity. Required thesuperficies and solidity of the tetraedron, whose linear edgeis 3 inches.

1 *73205 x 3 a = 15 *588 for its superficies,

0*11785 x 3 8 =3*18195 for its solidity.

llexaedron is the same thing as the cube, already described.

Octaedron. — Multiply the square of its side by the diagonal,and one third of the product will give the solid content.

Required the superficies and solidity of an octaedron, A, the linearsides of which at B E C is each 2 inches : taking from the table3*46410 x 2 2 = 13*85640 for the superficial content,

0*47140 x 2 2 = 3*77 1 20 for its solidity.

Dodecaedron .—Multiply the content of one of its pyramids, E, by 12,because the figure is composed of twelve equal pyramids having a regularpentagon for the base, their summits being the centre of the dodecaedron.

Required the superficies and solid content of a dodecaedron, whoselinear edges are 2 inches.

20*64573 x 2 2 = 82*58292 for the superficies,

7*66312 x 2 a =61 *30466 for its solidity.

Icosaedron. — Multiply the content of one of its pyramids, N, by 20,because it contains twenty equal pyramids having equilateral trianglesfor bases, and their summits in the centre of the body.

Required the superficies and solid content of the icosaedron N.

8*66025 x 2* = 34*64100 for the superficies,

2*18169 x 2 a = 17*43352 for the solidity.

The side of any of the five Platonic bodies being given, to find thediameter of a sphere that may either be inscribed in that body or cir-cumscribed alxiut it, or that is equal to it: As the respective number inthe following table, under the title inscribed, circumscribed or equal isto 1, so is the side of the given Platonic body to the diameter of itsinscribed, circumscribed, or equal sphere.

The side of any one of the five Platonic bodies being given, to find theside of the other four bodies that may be equal in solidity to that of thegiven body : As the number under the title equal, in the third column ofthe second table, which stands against the given Platonic l>ody, is to thenumber under the same title against the body whose side is sought, so isthe side of the given Tlatonic body to the side of the body sought.

Fig. 1216

Fig. 1217.

Fig.1219.

Fig. 1220.

Fig 1221.

t