7. A Right Prism is one whose other faces are perpendicular to its bases. (Corollary.) Its lateral faces are rectangles. 8. A prism is called triangular, quadrangular, or pentagonal, according as its base is a triangle, a quadrangle, or a pentagon; and so on. 9. A Parallelopiped is a prism whose bases are parallelograms. (Corollary.) It follows that all its faces are parallelograms. 10. A Right Parallelopiped has all its faces rectangles. 11. A Cube is a parallelopiped whose faces are all squares. (Corollary.) It follows that its faces are all equal, and the parallelopiped right. 12. A Cylinder is a right prism whose parallel faces are regular polygons of an infinite number of sides, that is, whose parallel faces are circles. A cylinder can be described by the revolution of a rectangle about one of its sides which remains fixed. The side opposite the fixed side describes the convex surface, and the other two sides the two circular bases. Thus the rectangle ABCD revolving about BC would describe the cylinder, the side A D the convex surface, and A B, DC the circular bases. 13. The Axis of a cylinder is the straight line joining the centres of the two bases; or it is the fixed side of the rectangle whose revolution describes the cylinder; as BC. THEOREM I. 14. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Let AH be a right prism; its convex surface is equal to FG+GH+HI+IK+K F A multiplied by its altitude A F B C E (D K For the convex surface is equal to the sum of the rectangles AG, BH, CI, &c. The area of the rectangle AGFG AF; the area of BHGHX BG; of CI=HIX CH; and so on. But the edges A F, BG, CH, &c. are equal to each other and to the altitude of the prism; and the bases of these rectangles together form the perimeter of the prism. Therefore the sum of these rectangles, that is, the convex surface of the right prism, is equal to the perimeter of its base multiplied by its altitude. 15. Corollary. As a cylinder is a right prism (12), this demonstration includes the cylinder. If, then, R= the radius of the base, and A= the altitude of a cylinder, the convex surface 2 RA. π THEOREM II. 16. The sections of a prism made by parallel planes are equal polygons. Let the prism A H be intersected by the parallel planes LN and QS; then LN and QS are equal polygons. For LM, MN, NO, &c. are respectively parallel to QR, RS, ST, &c. L (IV. 9), and similarly situated; therefore the angles L, M, N, O, P are respectively equal to the angles Q, R, S, T, U (IV. 11); and the polygons LN and QS are mutually equiangular. Also the sides LM, MN, NO, &c. are Q P دم. N S T H respectively equal to QR, RS, ST, &c. (I. 62). Therefore the polygons, being mutually equiangular and equilateral, are equal (II. 6). 17. Cor. 1. A section made by a plane parallel to the base is equal to the base. 18. Cor. 2. A section of a cylinder made by a plane parallel to the base is a circle equal to the base. THEOREM III. 19. Prisms having equivalent bases and equal altitudes are equivalent. Let AC and FH be two prisms having equal altitudes and their bases BC, G H equivalent; the prisms are equivalent. A F D Let DE and IK be sections made by planes respectively parallel to the bases BC and GH; these sections are respectively equal to the bases (17); there- B fore the section DE is equivalent to IK, at whatever distance from K E H the base either may be. If, therefore, the planes of these sections move, remaining always parallel to the bases, as the sections will always be equivalent, it is evident that in moving over an equal length of altitude the sections will move over equal volumes; therefore, as the altitudes are equal, the prisms are equivalent. 20. Corollary. Any prism is therefore equivalent to a right prism having an equivalent base and an equal altitude. THEOREM IV. 21. The volume of a right parallelopiped is equal to the product of its three dimensions. Let A D be the right parallelopiped; then its volume is equal to BCX BEX BA. Suppose BF, the A D linear unit, is contained in BC four times, in BE five times, and in BA seven times; then dividing BC, BE, BA respectively into four, five, and seven equal parts, and passing planes through the several points of division parallel to the sides of the parallelopiped, there will be formed a number of cubes equal to each other (19), and each equal to the cube whose edge is the linear unit. It is evident also that the whole number of cubes is equal to the product of the three dimensions, or 4 × 5 × 7 = 140. This demonstration is applicable, whatever the number of units in the linear dimensions may be. Therefore the volume of a right parallelopiped is equal to the product of its three dimensions. B F E 22. Scholium. If the three dimensions are incommensurable, the linear unit can be taken infinitely small, that is, so small that the remainder will be infinitesimal and can be neglected. 23. Cor. 1. As the base is equal to BCX BE, the volume of a right parallelopiped is equal to the product of its base by its altitude. 24. Cor. 2. The volume of a cube is equal to the cube of its edge. 4* THEOREM V. 25. The volume of any prism is equal to the product of its base by its altitude. For any prism is equivalent to a right parallelopiped, having an equivalent base and the same altitude (20); and the volume of the equivalent right parallelopiped is equal to the product of its base by its altitude; therefore the volume of any prism is equal to the product of its base by its altitude. 26. Corollary. As a cylinder is a right prism, this demonstration includes the cylinder. If, therefore, R = the radius of base, A = the altitude, and the volume of a cylinder, 27. Similar prisms are as the cubes of their homologous lines. Let AD and EH be similar prisms whose altitudes are IK and MN. Let V represent the volume of A D, and v the volume of EH; then V: v=IK3 : M N3 — AC3: EG3 For (25) VCDX IK and v = G H × MN, therefore CDXIK: GHXM N CD: GH A B EM F DG H K P CO2: G P2 V: v But (II. 31) and (4) IK: MN: = CO: GP Multiplying the last two proportions together we have CD × IK: GH × MN: therefore (Pn. 11) = C 03: G P3 V: v=CO3 : G P3 But in similar solids homologous lines have a constant ratio (4); therefore V: v as the cubes of any homologous lines. |