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But, in the right-angled triangle ABH, we have,
AH AB cos A AB cos
(Trig., Th. 5).
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 2n
25. To determine the angle included between two adjacent faces of either of the regular polyedrons, let us suppose a plane to be passed perpendicular to the axis of a polyedral angle, and through the vertices of the polyedral 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 polyedral angles, and each side will be a diagonal of one of the equal faces of the polyedron.
Let D be the vertex of a polyedral 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, and suppose AE, BE, to be the lines in 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.
Then, if we represent by n the number of faces which meet at the vertex of the solid angle, and by m the number of sides of each face, we shall have, from what has already been shown,
BF BC cos
and EB BC sin
sin FEB = sin A, to the radius of unity;
This formula gives, for the diedral angle formed by
any two adjacent faces of the
Having thus found the diedral 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 REGULAR POLYEDRONS WHOSE EDGES ARE 1.
26. To find the solidity of a regular polyedron.
1. 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.
Or, 2. 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, having
their common vertex at the centre of the polyedron. 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 (B. II., P. 10); that is, as the cubes of their homologous edges (B. VII., P. 20); that is, as the cubes of the edges of the polyedron.
Ex. 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 edge is
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 in stead 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.