Let ABCDE represent a regular spherical polygon, and let P be the pole of a small circle passing through its vertices. Suppose P to be connected

with each of the vertices by arcs of great circles; there will thus be formed as many equal isosceles triangles as the polygon has sides, the vertical angle in each being equal to 360° divided by the number of sides. Through P draw. PQ perpendicular to AB: then will AQ

E

D

P

B

be equal to BQ. If we denote the number of sides by n,

the angle APQ will be equal to

360°
2n

180°

or

n

In the right-angled spherical triangle APQ, we know the base AQ, and the vertical angle APQ; hence, by Napier's rules for circular parts, we have,

sin (90°- APQ) cos (90° - PAQ) cos AQ;

or, by reduction, denoting the side AB by 8, and the angle PAB, by A,

To find the volume of a regular polyedron.

126. If planes be passed through the centre of the polyedron and each of the edges, they will divide the polyedron into as many equal right pyramids as the polyedron has faces. The common vertex of these pyramids will be at the centre of the polyedron, their bases will be the faces of the polyedron, and their lateral faces will bisect the diedral angles of the polyedron. The volume of each pyramid will be equal to its base into one-third of its altitude, and this multiplied by the number of faces, will be the volume of the polyedron.

It only remains to deduce a formula for finding the distance from the centre to one face of the polyedron.

Conceive a perpendicular to be drawn from the centre of the polyedron to one face; the foot of this perpendicular From the foot of this per

will be the centre of the face. pendicular, draw a perpendicular to either side of the face in which it lies, and connect the point thus determined with the centre of the polyedron. There will thus be formed a right-angled triangle, whose base is the apothem of the face, whose angle at the base is half the diedral angle of the polyedron, and whose altitude is the required altitude of the pyramid, or in other words, the radius of the inscribed sphere.

Denoting the perpendicular by P, the base by b, and the diedral angle by A, we have Formula (3), Art. 37, Trig.,

P = b tan A;

bat bis the apothem of one face; if, therefore, we denote the number of sides in that face by n, and the length of each side by s, we shall have (Art. 101, Mens.),

The volumes of all

hence, the volume may be computed.

the regular polyedrons have been computed on the supposition that their edges are each equal to 1, and the results are given in the following

From the principles demonstrated in Book VII., we may write the following

RULE.

To find the volume of any regular polyedron, multiply the cube of its edge by the corresponding tabular volume ; the product will be the volume required.

EXAMPLES.

1. What is the volume of a tetraedron, whose edge is 15? Ans. 397.75.

What is the volume of a hexaedron, whose edge is 12 ?
Ans. 1728.

3. What is the volume of a octaedron, whose edge is 20? Ans. 3771.236.

4. What is the volume of a dodecaedron, whose edge is 25 ? Ans. 119736.2328.

5. What is the volume of an icosaedron, whose edge is 20 ?

Ans. 17453.56.

REMARK. In the following table, in the nine right hand columns of each page, where the first or leading figures change from 9's to O's, points or dots are introduced instead of the O's, to catch the eye, and to indicate that from thence the two figures of the Logarithm to be taken from the second column, stand in the next line below.