CHAMBERS'S INFORMATION FOR THE PEOPLE. 



AB = ab X cir. of which OH is radius, 



BC = be X cir. of which OH is radius, 



CD = cd X cir. of which OH is radius. 



Taking the sum, we have surface generated by 



AB + BC + CD = (ab + be + cd} X cir., 



of which OH is radius. 



This being true, whatever may be the number 

 of chords AB, BC, &c., let that number be inde- 

 finitely increased, therefore 



surface of zone = ad X cir. of sphere. 

 Let s denote the surface of zone, whose altitude 

 is A, the radius being r ; .'. s 2irrh. 



Let h = 2r t and we have for the total surface of 

 the sphere S = 4^^, 



or the surface of the sphere is equal to four times 

 the area of a great circle. 



OF VOLUMES. 

 The Parallelepiped Cube. 



PROP. XVI. The volume of a rectangular 

 parallelepiped is equal to the product of its three 

 dimensions. 



Thus let the side B3 in the figure contain 5 

 units, AB 4, and AE 3 ; 

 then it is evident that we 

 may consider the whole 

 solid as being made up of 

 slices AB, bf t cC, and we 

 have one such slice for 

 every unit in AE that is, 

 3. Now in every slice we 

 have as many piles as 

 there are units in AB 

 that is, 4 ; and each pile 

 has as many cubes as there 

 are units in Bt>, so that we have in the figure 3 

 slices, or 3 X 4 piles, or 3 X 4 X 5 cubes. The 

 volume of the cube is evidently found by taking 

 the cube of the side. 



The Prism. 



PROP. XVII. The volume of any prism is 

 equal to the product of its 

 base by its altitude. 



(i.) Let ABC A' be 

 a triangular prism. This 

 prism is equivalent to 

 one-half of the parallele- 

 piped ABCD A' con- 

 structed upon the edges 



AB, BC, BB', and it has *(:--/- -;D 



the same altitude. The 

 volume of the parallele- 

 piped is equal to its base 

 multiplied by its altitude ; 

 therefore the volume of the triangular prism is 

 equal to its base ABC, the half 

 of BD, multiplied by its alti- 

 tude. 



(2.) Let ABCDE A' be 

 any prism. It may be divided 

 into triangular prisms by planes 

 passed through a lateral edge 

 AA' and the several diagonals 

 of the base. The volume of 

 the given prism is the sum of 

 the volumes of the triangular 

 prisms, or the sum of their 



638 



bases multiplied by their common altitude. Bi 

 the sum of- the bases is the base ABCD ; ther 

 fore 



volume of prism = area of base X altitude. 



The Cylinder. 



PROP. XVIII. The volume of a right cylinder 

 on circular base is equal to the product of the base 

 by the altitude. 



Inscribe within the cylin- 

 der the right prism ABCD 

 ...HA/...H'. The volume 

 of this prism is equal to the 

 area of the base multiplied 

 by the height. Conceive the 

 number of faces of the in- 

 scribed prism to augment 

 continually, then the volume 

 of the prism approaches 

 nearer and nearer to that of 

 the cylinder, and ultimately 

 coincides with it ; the area of 

 the base of the prism ap- 

 proaches nearer and nearer 

 to that of the base of the cylinder, and the height 

 of the prism is always that of the cylinder ; hence 



volume of cylinder = area of base X height. 



The Pyramid. 



PROP. XIX. To find the volume of a triangu- 

 lar pyramid. 



Let S ABC be a triangular pyramid. Through 

 one edge of the base, as AC, pass a plane ACDE 

 parallel to the opposite lateral edge SB, and 

 through S pass a plane SED parallel to the base ; 

 the prism ABC E has the 

 same altitude as the given 

 pyramid. Taking away the 

 given pyramid S ABC from 

 the prism, there remains a 

 quadrangular pyramid, whose 

 base is the parallelogram 

 ACDE and vertex S. The 

 plane SEC, passed through 

 SE and SC, divides this pyra- 

 mid into two triangular pyra- 

 mids S AEC and S ECD, 

 which are equivalent to each 

 other, since their triangular bases AEC, ECD are 

 halves of the parallelogram ACDE, and their 

 common altitude is the perpendicular from S 

 upon the plane ACDE. The pyramid S - ECD 

 may be regarded as having ESD for its base, and 

 its vertex at C, therefore it is equivalent to the 

 given pyramid S ABC ; so also will the pyra- 

 mid S - EAC be equal to S ABC ; therefore 

 the given pyramid is equal to one-third of the 

 whole prism, or 



volume of pyramid = \ volume of prism. 



The Cone. 



PROP. XX. To find the volume of a cone. 



Inscribe within the cone SAD the regular pyra- 

 mid SABCDEF ; the volume of this pyramid is 



