B E To find the cubical content of any prism : RULE.—Multiply the area of either of the equal ends or bases, by its perpendicular distance from the opposite end, that is, by the altitude of the figure. Example (1.) The length, C D, of a parallelo- A piped is 5 inches, the breadth B C, 3 inches, and the depth A B, 2 inches. Required the solidity. A B X B C = 3 x 2 = 6; 6 x 5 (CD) = 30. Answer. 30 cubic inches. Each of the surfaces being divided into squares by parallel lines one inch apart, as shown in the diagram, it is seen that the end surface A C contains 6 square inches, and that the solid portion A E contains 6 cubic inches; but the entire solid being made up of 5 such portions as A E, the content of the whole must be 6 x 5 = 30 cubic inches." As similar reasoning may be applied to every solid of this form, where there are two equal and similar ends connected by parallel side-faces, the generality of the rule above given will be evident. In that particular case of the prism called the parallelopiped, as illustrated by the present diagram, the rule for finding the content is usually worded as follows:-Multiply together the numbers expressing the length, breadth, and height (or depth) of the gure. This process is the same in effect as that directed by the rule here laid down, but it has the disadvantage of applying to one description of prism only. Example (2.) What is the content of a triangular prism (see fig. 1, last diagram but one) the height of which, A C, is 10 feet, the three sides of the triangle forming its base being respectively, 5, 4, and 3 fect ? Half sum of the three sides of the triangle C D E = 6. 6 - 5 = 1 6 = 3; 6 x 3 x 2 x1= 36; 36,= 6, Area of triang. C D E. And 6 x 10 (height of prism) = 60 cubic feet. Answer. Example (3.) Required the solidity of a pentagonal prism, (see fig. 2, last diagram but one) where each side of the base measures 7.5 inches and the height 12.8 inches. Tab, no. (pa. 181.) Perimeter, (68S2 x 7.5 x 37.5) x = 96.778 96.778 x 12.8 = 1238.76. Answer. 1238.76 cub. inches. Since an oblique parallelogram has the same area as a rectangle on the same base and of equal altitude (Euclid I., 35), so, it is manifest, an oblique prism will have the same solidity as a rectangular prism on the same base and of the same altitude; for the volume of every prism may be considered to be made up of a number of plates of indefinite thinness piled one upon the other; and the number of such plates composing prisms of equal altitudes will evidently be the same, when the component plates of each have the same thickness. This being the case, the perpendicular height, and consequently, the volume of the prism must be the same whether the plates be piled obliquely or otherwise. The truth of this property is very simply verified by arranging a pack of cards in a perpendicular heap, or with the edges upright, and then causing the cards to lean towards the end of the pack. When thus placed in a slanting position, the cards will still form a prism having the same base and altitude as before. A Pyramid is a solid, the base of which is any plane rectilinear figure, a triangle, a square, &c., and the side-faces of which are triangles, all terminating in ono common vertex. (See figs. opposite.) Pyramids, like prisms, take their names from the forms of their bases. Thus, when the base is a squaro, the solid is called a square pyramid; when the base is a triangle, a triangular pyramid, and so on. If the perpendicular from the vertex falls on the centre of the base, the solid is said to be a right pyramid, and when otherwise, an oblique pyramid. = 440. 3 2 (675) To find the solidity of any pyramid.-RULE.—Multiply the area of the base by the altitude (perpendicular height) of the figure ; one-third of the product will be the solid content. Example (1.) What is the solidity of a rectangular pyramid, two adjacent sides of the base measuring 11 and 8 inches respectively, and the altitude 15 inches? 11 x 8 = 88, Area of base ; 88 x 15 Answer. 440 cubic inches. Example (2). The base of a pyramid is an equilateral triangle, each side of which is 6.75 feet; the altitude is 32 feet, 4 inches. Required the solid content. By the Rule for equilateral triangles on page 177, the area of the base = 19.728 x 321 x 1.732 = 19.728 square feet. 3 = 212.624. 3 Answer. 212.624 cubic feet. The following is perhaps the most simple illustration that can be given of the truth of the Rule for computing the solidities of pyramids. Let D E be a cube, D B being one of its six equal square faces. Let G be the middle or centre point of the cube. It is evident that by means of lines drawn to G from A, B, E, F, the four angular points of one of the faces of the cube, we may form a square pyramid; it is evident also that the cube consists of six such pyramids; and, therefore, the solidity of each pyramid will be a sixth of that of the cube. But the solidity of the cube is the product of the base A B E F by the height BC; therefore, the solidity of the pyramid is the product of the same base by one-sixth of B C, that is, by one-third of half B C, or one-third of the height of the pyramid. Now, it is proved in geometry, (Euclid XII., 7) that, as the area A of any plane triangle is the product of the linear base by half the B perpendicular altitude, so the solidity of any pyramid is the product of the superficial base by a third of the perpendicular altitude. The solidity of a pyramid is, in fact, one-third of that of a prism having the same base and altitude. A segment of a solid is any portion cut off from the top by a plano parallel to the base. Thus C D n E is a segment of the pyramid C B A. A frustum of a solid is the portion that remains after a seg. ment is removed. A D E B is a frustum of the pyramid C B A.* The term frustum is used chiefly with regard to those solids, which in their complete state, taper to an edge or a point, namely, the pyramid, the cone, and the wedge. It will be evident, that in the case of a prism, (and also in the case of a cylinder, as may be seen on referring to the B drawing of that figure, a few pages in advance,) any section made by a plane parallel to the base can only divide the solid into a smaller prism or cylinder similar to the original. No distinguishing name, therefore, is required for the part so cut off. с a D To find the content of a frustum of a pyramid.-RULE.-Add together the areas of the two ends or bases, and the square-root of the product of these areas, (i.e., the mean proportional between them). Then multiply the sum by one-third of the altitude of the frustum. Example. In A E a frustum of the square pyramid C B A, let B m, an edge of the greater end = 7 inches, E n, an edge of the smaller end 4 inches, and the perpendicular height, B E, = 8 inches. What is the solidity of the frustum ? а • It is customary to apply the term frustum also to any part of such solids as the spheroid, paraboloid, &c., which is intercepted between two planes passing through the figure at any distance apart, these planes being either parallel or inclined to each other. 7 x 7 49, area of the end A B. 7 x 4 28, mean proportional between 49 and 16. IIence, (49 + 16 + 28) X = 93 x 2z = 248. Answer. 218 cub. inches. The reasoning by which the truth of this compendious rule is established would harilly be intelligible to persons who have no knowlege of elementary algebra. It must here be sufficient to state generally, that any frustum of a pyramid is equal to the sum of three pyramids of the same altitude, whose bases are:-1st, the lower base or end of the frustum: 2001 , the upper base; and 3rd, a surface which is a mean proportional between the two. Hence, the solid content of such a frustum, whatever be the figure of the base, may be found by this rule, -Add into one num the areas of the two ends and the square-root of their product; one-third of that sum will be the area of the base of an equal prism having the same altitude as the frustum, and consequently, this area multiplied by the altitudo will give the solidity of the frustum. That which would suggest itself as the most simple and obvious mode of ascertaining the content of a pyramidal frustum, is to consider the frustum as equal to the differenco of two pyramids--the entire pyramid of which it forms the lower portion, and the smaller pyramid cut off at top. In order to obtain the date necessary for computing the solidities of these pyramids we have only to proceed as follows :-As C m is parallel to B m (last fig.) the triangles C E n and C B m are similar. Hence by a corollary, easily deducible from Euclid VI., 4, B m : En :: BC:EC. But BC EC + B E (the given height of the frustum); and since in a proportion the product of the extremes is equal to the product of the midille terms (ARITHMETIC, page 107), we have B m X EC = En X EC + En x B E. From this it appears that to tind EC we should multiply the altitude of the frustum by a side of the upper end, and divide the product by the difference between a side of the upper and a side of the lower end. The heights of both pyramids being thus determined, and the areas of their ends or bases being known, their volumes can be computed and the difference of the results taken, which will represent the content of the frustum. For instance, by the rule just derived, and retaining the dimensions in the last 8 X 4 32 example, the height E C proves to be = and the area of the surface 3 3 32 EDX one-third of = 56-89 = content of upper pyramid. Also content of entire pyramid = $ x 56 x 49 = 304.89; and 304•89 - 56.89 = 248, as before. But it is evident that this process is not so convenient as that directed by the Rule, although the latter indicates merely a more compact manner of performing the same operations. It is important to observe, that since all plane rectilinear figures admit of having their areas resolved into as many triangles as they have sides, by taking any point within them as the common vertex of the component triangles (See page 179) so all solids whatover admit of having their volumes resolved into as many pyramids as they have surfaces, by taking within their volumes any point as the common vertex of the component pyramids, and drawing lines from that point to their several angles, which lines will, it is evident, form the edges of the triangular faces of the component pyramids. The species of pyramids into which the solid is thus resolved will depend on the kind of figures formed by the faces of the solid ; but since all pyramids whatever can be resolved into triangular pyramids by drawing planes through their vertices and the diagonals of their bases, it follows that all solids whatever, having plane faces bounded by straight edges, admit of being ultimately divided into triangular pyramids.* As an example of the application of this general property of rectilinear solids, suppose a central point to be taken within the frustum D B (last tig.) in a line with the middle of the altitude B E, and straight lines to be drawn from this point to the several angles of the surfaces A B and D Е, thus forming two square pyramids having these surfaces respectively as their bases. Also suppose that on each of the side-faces I m, n B, &c., a pyramid # Treatise on Geometry, Cab. Cycl. 2 E F is constructed having its vertex at the same central point within the figure. There would thus be in all six component pyramids, two with square bases and four with bases of the form of trapezoids; and it is manifest that these pyramids would together exactly make up the space enclosed by the surfaces of the frustum D B. Now to calculate the contents of these pyramids, we have 1st, the areas of the two squares A B and D E multiplied by one-third of half the perpendicular height B E, that is, by one-sixth of B E; and 2nd, as the distance of the common vertex of the pyramids from each of the side-faces D m, &c., must be equal to half the length of a line drawn across the middle of any one of the faces, and as this length must, from the uniform inclination of the edges B E, m n, &c., be the arithmetical mean between the pairs of parallels, B m, En, &c., we have the areas of the four trapezoids, D m, n B, &c., multiplied by onethird of one-half, or one-sixth of the altitude represented by the mean distance in question. The content of the frustum D B may therefore be computed as follows:Surface A B= 49 Surface Em = En + B m X B E (pa. 178) = \ x 8 = 44 Surface DE= 16 Sum 4 65 x no. of equal surfaces X by dth of B E = 1} 176 Product...86.67 X fth of arith, mean between En and Bm= 11 Product = 161.33 Sum of areas of square pyramids = 86.67 trapezoidal , 161.33 Content of frustum... 248.00 By a similar process, adapted to the form of the bases of the component pyramids, the content of any rectilinear solid whatever may be determined, so that, except for the labour of the various calculations, no other rule is absolutely needed. A Wedge is a solid of which the base is a parallelogram or a trapezoid, but usually a rectangle ; two opposite sides meet in an edge, and the two ends are triangles whose vertices meet the extremities of that edge. In each of the accompanying diagrams, A F D B represents a wedge, the base being D B and the edge E F. When the lengths of the edge and base are equal, as in the first diagram, the wedge is a triangular prism. And it is manifest, from the equality of the pyramids E A K and F B I, that the oblique prism G F KH is equal to the right prism A F D B; so that the volume of the oblique prism is equal to the product of the area of ADE by the length A B, or is equal to the breadth K G x ļ the perpendicular height of the prism x the length G H. Let A B, the length of the base, be, as in the second of the preceding figures, greater than the length of the edge E F. The wedge may then be divided into two solids, viz., a triangular prism F G C, and a rectangular pyramid E D G. Now, if the given dimensions are, A B = 16, BC 2.25, E F = 10.5, and the height = 7, then the solidity of the prism is į of (7 * 2.25) x 10.5, and that of the pyramid is of (5.5 x 2.25 x 7), and the sum of these, viz., the solidity of the wedge = 111.5625. Again, suppose A B less than E F; then by making G B equal to E F, &c., the wedge will be the difference between the triangular prism F G C and the rectangular pyramid E DG. So that if A B is 10.5, B C 2.25, E F 16, and the alti. tude 7, the solidity of the prism, 1 x (7 x 2.25) x 16, diminished by that of the pyramid, { x (5.5 x 2.25 x 7), gives as the solidity of the wedge, 97.125. D A A с B But in practice it is more convenient to calculate the solidity of a wedge by the following method : 2 2 To find the solidity of a wedge.—Rule.—To the length of the edge add twice the length of the base, and take a third of the sum for a mean length; then multiply this mean length by half the product of the breadth and height. Example. Find the solidity of a wedge the height of which is 7 inches, the length of its back, or base, being 16 inches, the breadth of the base 2.25 inches, and the length of the edge 10.5 inches. } of (10.5 + 32) × 1 of (2.25 + 7) 111.5625 cubic inches, as before. The following is the most simple demonstration which can be given of this rule :-If the height be called h, the length of the edge e, the length of the base l, and the breadth of the base 6, then, when the base is longer than the edge, we have, according to the rules previously given, the solidity of the prism (hxb) x e, and that of the pyramid } (l—e) x B x h; the sum of these is šle + 2 l) x 2 5 h, which is the Rule. Also, when the edge is longer than the base, we have the solidity of the completed prism (h x b) x e, and that of the added pyramid } (e — 1) x bh; and, subtracting the latter of these from the former, we have, as before, } (e + 2 l) x 1 bh. (Hunter's Mensuration.—Gleig's Series.) A frustum of a wedge, usually called a Prismoid, is the portion remaining after a smaller wedge has been cut off by a plane parallel to the base. Thus, in each of the opposite figures, H DBF is a prismoid. This solid evidently consists of two wedges, as presented in the second á figure: the back, or base, of one being B D, and its edge H E; the base of the other G E, and its edge B C. Accordingly, the sum of the solidities of these two wedges will be the solidity of the prismoid. Suppose, in the second figure, A D is 6, A B, 5, I E, 7.5, G H, 3.5, and the perpendicular height of each wedge, 4.5. The solidities of the two wedges are as follows : H D BE, of (7.5 + 12) x of (5 x 4.5) = 73.125 2 = 55.125 Therefore, the solidity of the prismoid is 128.25. But the cubic measure of a prismoid is usually calculated by the following more convenient method, called the Prismoidal Formula. H E B re D A : 3 To find the solidity of a prismoid.—RULE.—To the sum of the areas of the two parallel ends add four times the area of the middle section parallel to these ends ; then multiply the sum by one-sixth of the height. It should be observed that half the sum of the lengths of the parallel ends is the length of the middle section, and half the sum of their breadths is the breadth of that section. The preceding rule is applicable to frusta of pyramids as well as to those of wedges.* It is thought unnecessary to give examples under this rule • A satisfactory demonstration of this rule cannot be given without the aid of algebra, but the correctness of the rule may casily be inferred from what has been stated respecting the solidities of frusta in general. (See page 208.) |