VOLUME OP A PRISMOID.] 



MATHEMATICS. MENSURATION. 



653 



A, 



(fc. 



fc.) 



Tlien adding these together, and writing 



*i + A + k,-M+*. -8 



we have 5V=4 L (S + 

 9 



- 2 (s 



s> + - 



If A 



.'. 5 V = (2Ai + As) S 

 area of base. 



.-. V = A. ^' "^ ^ 2 "^" ^ 



+ s 



A. S. 



s> 



The student can easily prove that a similar formula 

 I is true in the case of a prism on a hexagonal base, or 

 indeed on any base which is a regular polygon, 



(10). To find the Volume of a Prismoid. 

 DEF. A prismoid is a solid of the form represented 



yi f . . 



in the accompanying 

 figure. A B C D is a 

 rectangle, and ABFE, 

 DEHG, are planes 

 perpendicular to the 

 planet of the rectangle, 

 mil C F are planes 

 ied at given angles 

 to the plane of the 

 rectangle, the lines E F B 

 and G H being parallel 

 to AB and CD. 



Through A D and B C 

 draw planes A D L K 



and B C N M perpen- 



dicular to the plane of the rectangle ; then divide the 

 given figure into three viz., AEKDLG, BDN, and 

 BMFCHN. 



6, M F = c N H = c, 



Then (o) the figure BDN is of the same kind as in 

 Article 8. Now the area D B = oA ; if therefore V t is 

 its volume, 



V, = (x + x + x, + x,) - ?* (x + x,) 



(b). Let V a be the volume of the figure AEKDLG, 

 (Fig. 44), draw planes through E D,K Fig. 44, 



and GD,K, dividing the figure into A 



three pyramids, DGL.K, AEK,D, 

 andDEG.K. 



The volume of OLD, K 



KL x,6, fc_fc 



Similarly, volume of A E K,D =- xb. 



o 



Now the volume of GED,K is to 

 that of A D E, K as G E D is to ADE, 

 or as A E to G D ; or since triangles 

 A E K, GDL are similar, these 

 volumes are as A K to DL, or as 

 EKtoGL. 



.'. Volume GED,K 



-. x5 



(c). Similarly, if V, is the volume of MBFCNH 

 (Fig. !.;, 



_h 

 3 ~ 6 



Now if V be the required volume, 



V=V, +V S +V,. 

 /. V = ^ j 3 a (x + *,} + x, (6, + c,) + x (b + c) 



or rearranging the right-hand side of the equation, 



V = 



) 



6, +e, 



Now in this expression, 

 * \ + ~T~") "* the aroa of A B F E ' ( Fis- i3 ^ 



DCHG - 



and (x + *,) 2a + -^-+ ^ t ^ it \ which equals 



U clearly four times the area of a section made by a 

 plane parallel to ABFE and DEHG, and half way 

 between them. 



Hence the solid content of a prismoid is found by the 

 following rule: "To the areas of the ends, add four 

 times the area of the mean section ; multiply this sum by 

 one-sixth of the height of the figure ; this product is the 

 volume required." 



In practical cases it will generally happen that 6 = c, 

 and 6, = c t . This does not affect the enunciation of the 

 rule, but simplifies the formula, which becomes 



(11). To find the Solid Content of a Railway Cutting. 



In the last article, the figure (Fig. 43) is very nearly 

 that of a portion of a railway cutting, in which A B C D 

 is the road, A G and B H, the sloping sides of the 

 embankment, and hence the solid content required can 

 be found by means of the rule given in the last article. 

 It is to be observed that the rule requires E G and F H 

 to be straight lines, or, as an approximation, to be very 

 nearly straight lines, or E H to be a plane, which is not 

 true if the cutting be a long one. For this case we 

 derive the following rule from the above formula. 



Let ABC represent a section made by a vertical plane 

 of the hill to be cut through, A B the level of the road, 

 and suppose sections of the cutting to be made by planes 

 perpendicular to A B, at equal distances along that line, 

 viz., at M, M 2 M 3 . . . M 2n + ,, let the terminal sections 

 at A and b be a and b, and the sections at P t M t P a M 2 



,p l p a p a /> 2 i. P 2 + i , the number of sections being 



odd, and let the common distance between the sections 

 h. Now by last article the volume of portion 



