MENSURATION 23 68. Surface of a Prism. Since the lines A"B", B"C , &c. (fig. 41), which make up the perimeter of the right section are all in one plane perpendicular to the parallel edges A A ", B B ", &c. , they are perpendicular to these edges and are therefore the altitudes of the parallelograms A B B" A ", B C C "B ", &c., respectively. The lateral surface of the prism is equal to the sum of these parallelo grams, and therefore =A A "xA"B" + B B -xB"C" + ...... = A A "(A"B" + B"C" + ...... ), since A A " = B B " = &c. ; or the lateral surface of any prism is equal to the perimeter 01 its right section multiplied by the length of the prism. If the prism be right, that is, if the faces be perpendicular to the base, then its lateral surface is equal to the perimeter of its base multiplied by its length. The whole surface of any prism is obtained by adding to the lateral surface the areas of its bases. 69. If the prism be regular, that is, if the bases be regular polygons, then area of base = a 2 x -^ cot ( 18, 7), where n is the number 4 n of sides each of length a, and therefore n ,180 , volume = a- x cot - - x h , 4 n where h is the altitude of the prism. Again, if the prism be right and regular, then its lateral surface = nah + 2a 2 x cot ----- . 4 n 70. Volume of a Pyramid. Let VABC (fig. 42) be for sim plicity a triangular pyramid. Divide VA into n equal portions, and through the points of section draw planes parallel to the base ABC, and through BC and through the inter sections of these planes with VBC draw planes parallel to VA. Let h denote the altitude of the pyramid, then the distance of the base of the ?" prism from the vertex V h and, if A denote the area of ABC, we A have base of r th prism _ r"h 2 1 _ r 2 ^ A n z A 2 n- since, by a well-known theorem in solid geometry, the areas of sections of a pyramid made by planes parallel to the base are proportional to the squares of their altitudes. Thus we have r 2 base of r th prism = A. and therefore n 2 its volume = A x ( 67) Therefore volume of whole pyramid + r- + . +n" or the volume of any pyramid is equal to one-third of the area of its base multiplied by its height. From this we see that pyramids on equal bases are to one another as their altitudes. If the pyramid be regular, that is, if its base be a regular polygon the perpendicular through whose centre passes through the vertex, its volume = i x a 2 x - cot x h . 4 n 71. Surface of a Regular Pyramid. The lateral surface of the regular pyramid VABCDEF (fig. 43) is equal to the sum of the areas of the n congruent triangles which make up the lateral surface of the pyramid. Now area of triangle VAB = |AB x VG ; hence whole lateral sur face = |/iAB VG = ^nal, where I is the slant height and a the length of the side of the base. Again, if V0 = ft = altitude of pyramid, we have therefore whole surface = base + lateral surface 180 /~ - + ia / - = a- x - cot 4 , a? . 180 + cot- 4 n 72. The Prismatoid. If we have a polyhedron whose bases are two polygons in parallel planes, the number of sides in each being the same or different, and if we so join the vertices of these bases that each line in order forms a triangle with the preceding line and one side of either base, the figure so formed is called a "prismatoid," and holds in stereometry a position similar to that of the trapezium in planimetry. To make the investiga tion of the volume of the prismatoid as simple as possible, we take the case where the lower base is a polygon of four and the upper one 01 three sides. Let ABCDEFG (fig. 44) be the pris matoid, of which ABC or Aj is the upper and DEFG or A 3 the lower base, and let HLM be the Fl 8- 44 - section equidistant from the bases. Take any point P in this section and join it to the corners of the prismatoid. We thus divide the polyhedron into two pyramids PABC and PDEFG, and a series of polyhedra of which CPDE may be taken as ; specimen. Let h be the altitude of the prismatoid, then ^li is the altitude of each of the pyramids PABC, PDEFG, and hence volume of PABC = AAj , and volume of PDEFG = &AA 3 . Again join PH, PL, and LD, then volume of CPDE = 2 volume of CPDL, since DE = 2HL, and volume of CPDL = 2 volume of CPHL, hence volume of CPDE = 4 volume of C PHL. Now volume of CPHL = )>h x area of HLP, and therefore volume of CPDE = ^ x area of HLP. Similarly the volume of every such polyhedron is %h x the area of its own portion of the middle section. Hence if A 2 denote the area of the middle section we have volume of prismatoid = JAAj + JAA a + -*/( A. 2 73. Volume of the Frustum of a Pyramid. Let A A "B B "C C " (fig. 45) be a frustum of the pyramid VA B C , and let A, and A 3 denote the areas of the ends A B C , A" B "C " respectively. Let VP = x = altitude of pyramid VA "B "C ", and let PQ = h = altitude of frustum.
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