The similarity of the solid thus constructed with the given solid will easily appear from the results of (133) et seq.

(141) It should be observed, that in this book we confine our investigations to what are called convex polyedrons, or those which have no solid angles whose vertices are presented inwards. The most distinct test of a convex figure, whether plane or solid, is, that its perimeter or surface can be only intersected by a right line in two points.

(142) DEF.-The solid figure included between two parallel planes intersecting a pyramid is called a truncated pyramid.

PROPOSITION XXV.

(143) The volume of a truncated pyramid is equal to the sum of the volumes of three pyramids having the same altitude as the truncated pyramid, and whose bases are those of the truncated pyramid and a mean proportional between them.

This proportion is most simply investigated by the aid of algebraical notation. Let a, a' be homologous sides of the similar bases (119) of the truncated pyramid. The squares of these lines are as the areas of the bases, or the square of each line bears the same ratio to the area of the base of which it is a side. Hence if m be such a number that ma2 is equal to the one base, m a2 will be equal to the other. Now, suppose the sides of the truncated pyramid produced till they meet so as to complete the pyramid, and let a perpendicular be drawn from the vertex to the parallel bases of the truncated pyramid. Let the length of this perpendicular drawn to the greater base be h, and to the lesser h. The volume of the pyramid having the greater base ma2 is then hm a2 (127), and that of the lesser h'm a'2, and therefore the volume of the truncated pyramid ishma fh'm a22 = m (ha2 - h'a2). But (122) h: h':: a: a', and hence we easily infer

Now let V be the volume of the truncated pyramid

But hh' is the altitude of the truncated pyramid. Call this H; and if a-a's be divided by a a', the quote is a2 + aa+a. Hence we have

or

aa',

Now ma is the greater base, and H m a2 is the volume of a pyramid whose base is ma2 and whose height is H, or that of the truncated pyramid. In like manner Hm a2 is the volume of a pyramid whose base is the lesser base ma' of the truncated pyramid, and whose height is H, that of the truncated pyramid; and, lastly, Hmaa' is the volume of a pyramid whose base is ma a' (a mean proportional between ma2 and ma'2) and whose altitude is H, that of the truncated pyramid.

PROPOSITION XXVI.

(144) The volume of a solid included by two planes intersecting the sides of a triangular prism, is equal to the sum of the volumes of three pyramids whose common base is either of the sections, and whose vertices are the vertices of the angles of the other section.

If the two intersecting planes be parallel, the solid is a triangular prism, and the three pyramids are equal, and hence the proposition becomes identical with (126).

b

a

If the planes be not parallel, let the triangular sections be A B C and abc. Draw the plane BC a. This divides the solid into the pyramid ABC u, and the solid a b c BC. Draw the plane ab C. This divides the latter solid into two pyramids, whose bases are b c C and B C b, and of which a is the common vertex. The pyramid whose base is B C b and vertex a is equal to that with the same base B Cb and vertex A; because a A is parallel to the plane B C b, and therefore the two pyramids have equal altitudes. But the pyramid A Ca B is that which is on the base ABC, and has the vertex a. Hence, in like manner, the pyramid Cacb is equal

B

A

to Cbc A, for they have the same base Cbc and equal altitudes, since A a is parallel to that base. But the pyramid Abc C is equal to the pyramid A B c C, for they have the same base Ac C and equal altitudes, since Bb is parallel to that base. Hence it appears that the three pyramids A B Ca, A Ccb, and Ba Cb, into which the solid is resolved by the planes Ba C and ab C through the vertex a, are respectively equal to three pyramids, of which A B C is the base, and whose vertices are at the points a, b, c.

BOOK III.

Of the Regular Solids.

(145) DEF.-A regular solid is one whose faces are equal regular polygons, and whose solid angles are equal.

PROPOSITION XXVII.

(146) There cannot be more than five regular solids.

1o. Let the faces be equilateral triangles. A solid angle may be formed by three, four, or five plane angles, each of which is two thirds of a right angle. But six or more angles of this magnitude would be equal to or greater than four right angles; and, consequently, could not form a solid angle. The number of regular solids, therefore, whose faces are triangular cannot exceed three.

2o. Let the faces be squares. A solid angle may be formed of three right angles, but not of a greater number. Wherefore there is but one regular solid with square faces.

3. Let the faces be pentagons. A solid angle may be formed of three angles of a regular pentagon; for the magnitude of one is six fifths of a right angle, and therefore the aggregate magnitude of three such angles is of a right angle, or three right angles and three fifths, which is less than four right angles. But four or more such angles will be evidently greater than four right angles, and therefore cannot form a solid angle. Hence there cannot be more than one regular solid with pentagonal faces.

4°. If the faces were hexagons the angles would be four thirds of a right angle, and three such angles would be equal to four right angles, and therefore could not form a solid angle; and it is evident that no greater number of such angles than three could form a solid angle. If the faces were polygons with more than six sides their angles would be greater than those of a

regular hexagon, and similar observations would be applicable. Hence no regular solid can have faces with more sides than five. Hence we infer

1o. That there cannot be more than five regular solids.

2o. That three of these have triangular faces, one square faces, and the remaining one pentagonal faces.

3°. That the solid angles of the three regular solids with triangular faces, are formed of three, four, and five plane angles, and that the solid angles of the others are formed of three plane angles.

PROPOSITION XXVIII.

(147) To construct a regular solid, whose faces are triangular, and whose solid angles are contained by three plane angles.

P

From the centre O of an equilateral triangle AC B draw a perpendicular OP to its plane, and on this perpendicular from the points A, B, C inflect lines A P, B P, CP equal to the side of the equilateral triangle. It is evident that these lines will meet the perpendicular in the same point P. A pyramid will thus be formed on the base A B C, having its vertex at P, the four faces of which will be equal equilateral triangles. This is therefore the regular solid required.

This solid is therefore the regular tetraedron.

(148) COR. 1.-It is evident that each pair of faces of the tetraedron form the same angle. The magnitude of this angle may be easily calculated by spherical geometry and trigonometry, (160). It is, in fact, the angle of an equilateral spherical triangle, whose side is 60°. The calculation is as follows. Let the angle be A, we have then

cos a

cos A sin b sin c cos b cos c = 0. See my Trigonometry (181), first formula of [1].

In this let a, b, c be supposed to be each = 60°. Since

Hence the angle under each pair of faces is one whose cosine is