3. What is the solidity of a spherical segment with one base, the diameter of the sphere being 20, and the altitude of the segment 9 feet? Ans. 1781.2872.

PROBLEM XIX.

To find the surface of a spherical triangle.

RULE.-1. Compute the surface of the sphere on which the triangle is formed, and divide it by 8; the quotient will be the surface of the tri-rectangular triangle.

2. Add the three angles together; from their sum subtract 180°, and divide the remainder by 90°: then multiply the trirectangular triangle by this quotient, and the product will be the surface of the triangle (Book IX. Prop. XX.).

1. Required the surface of a triangle described on a sphere whose diameter is 30 feet, the angles being 140°, 92°, and 68°. Ans. 471.24 sq. ft.

2. Required the surface of a triangle described on a sphere of 20 feet diameter, the angles being 120° each. Ans. 314.16 sq. ft.

PROBLEM XX.

To find the surface of a spherical polygon.

RULE.-1. Find the tri-rectangular triangle, as before.

2. From the sum of all the angles take the product of two right angles by the number of sides less two. Divide the remainder by 90°, and multiply the tri-rectangular triangle by the quotient: the product will be the surface of the polygon (Book IX. Prop. XXI.).

1. What is the surface of a polygon of seven sides, described on a sphere whose diameter is 17 feet, the sum of the angles being 1080° ? Ans. 226.98. 2. What is the surface of a regular polygon of eight sides. described on a sphere whose diameter is 30, each angle of the polygon being 140°? Ans. 157.08

OF THE REGULAR POLYEDRONS.

In determining the solidities of the regular polyedrons, it Decomes necessary to know, for each of them, the angle contained between any two of the adjacent faces. The determi nation of this angle involves the following property of a regu lar polygon, viz.

Half the diagonal which joins the extremities of two adjacent sides of a regular polygon, is equal to the side of the polygon multiplied by the cosine of the angle which is obtained by dividing 360° by twice the number of sides: the radius being equal to unity.

But in the right-angled triangle ABH, we have

F

H

AH AB cos A=AB cos (Trig. Th. I. Cor.)

360°
2n

REMARK 1.-When the polygon in question is the equilateral triangle, the diagonal becomes a side, and consequently half the diagonal becomes half a side of the triangle.

REMARK 2.-The perpendicular BH=AB sin Th. I. Cor.).

360°

2n

(Trig.

To determine the angle included between the two adjacent faces of either of the regular polyedrons, let us suppose a plane to be passed perpendicular to the axis of a solid angle, and through the vertices of the solid angles which lie adjacent. This plane will intersect the convex surface of the polyedron in a regular polygon; the number of sides of this polygon will be equal to the number of planes which meet at the vertex of either of the solid angles, and each side will be a diagonal of one of the equal faces of the polyedron.

Let D be the vertex of a solid angle, CD the intersection of two adjacent faces, and ABC the section made in the convex surface of the polyedron by a plane perpendicular to the axis through D.

Through AB let a plane be drawn perpendicular to CD, produced if necessary, A and suppose AE, BE, to be the lines in

F

B

which this plane intersects the adjacent faces. Then will AEB be the angle included between the adjacent faces, and FEB will be half that angle, which we will represent by A.

A

F

B

Then, if we represent by n the number of faces which meet at the vertex of he solid angle, and by m the number of sides of each face. we shall have, from what has already been shown,

BF=BC cos

360°
2n

and EB-BC sin

360°
2m

•

=sin FEB-sin A, to the radius of unity;

EB

hence,

This formula gives, for the plane angle formed by every two adjacent faces of the

Having thus found the angle included between the adjacent faces, we can easily calculate the perpendicular let fall from the centre of the polyedron on one of its faces, when the faces themselves are known.

The following table shows the solidities and surfaces of the regular polyedrons, when the edges are equal to 1.

A TABLE OF THE REGULAR POLYEDRONS WHOSE EDGES ARE 1.

PROBLEM XXI.

To find the solidity of a regular polyedron.

RULE I.-Multiply the surface by one-third of the perpendicular let fall from the centre on one of the faces, and the product will be the solidity.

RULE II.

-Multiply the cube of one of the edges by the solidity of a similar polyedron, whose edge is 1.

The first rule results from the division of the polyedron into as many equal pyramids as it has faces. The second is proved by considering that two regular polyedrons having the same number of faces may be divided into an equal number of similar pyramids, and that the sum of the pyramids which make up one of the polyedrons will be to the sum of the pyramids which make up the other polyedron, as a pyramid of the first sum to a pyramid of the second (Book II. Prop. X.); that is, as the cubes of their homologous edges (Book VII. Prop. XX.); that is, as the cubes of the edges of the polyedron.

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

2. What is the solidity of a hexaedron whose edge is 12? Ans. 1728.

3. What is the solidity of a octaedron whose edge is 20? Ans. 3771.236. 4. What is the solidity of a dodecaedron whose edge is 25? Ans. 119736.2328. 5. What is the solidity of an icosaedron whose side is 20? Ans. 17453.56.