CONTENTS OF SOLIDS.] 



MATHEMATICS. MENSURATION. 



or OB. 



If we suppose the quadrant to revolve round 

 Fig. 27. A O it will describe a 



hemisphere ; and if we 

 suppose a number of 

 i equal chords, Ap,, p-,' 



Pi, Pt, Pa, <tc ->. * be 

 drawn from point to 

 ! point of A B, these 

 chords in the revolution 

 will describe frustums of 

 cones : now the arc A R 

 is the limit of the chords 



ri * I r * ' f * * *> ' 



. . . and hence the area 

 of the portion of sphere 

 described by A R will be the limit of the sum of the 

 frustums of cones described by Ap, p l p z p$ p x . . . 



Let P Q (Fig. 28) be one of these chords : draw P M, 

 Q N perpendicular to A O. Draw O perpendicular to P Q, 

 and tn perpendicular to A O. Now tn is the radius of the 

 mean section of the frustum of 

 _A cone described by PQ, and 

 - u therefore area of that frustum 



Now tn = Ot sin. <OA. 



and PQ sin. PQN = NM. 



or since PQN = <OA. 



PQ sin. <OA = NM 



.-. tn. PQ = Ot. NM. 



And area of frustum 

 )(. NM. 



In Fig. 27, draw p. n t p 2 n- 2 p a n s perpendicularly to 

 A O. Then, since, Ap l p, p 2 p, p a are all equals, the 

 perpendiculars on them are equal, and therefore the sum 

 of areas of frustum of cone 



= 2 r Ot X (An, + n, n g + n 2 n 3 + ---- ) 

 = 2irO< x AK 



if we only consider the portion of sphere described by 

 A R. This is true, however great the number of chords, 

 and is therefore true in the limit ; but in the limit Ot 

 radius of sphere. 

 .". Area of portion of sphere, whose height is AN. 



= 2irOA. AN. 

 If we take the whole sphere. A N = 2 O A. 



.'. Area of sphere = 4>r. (O A) 1 - 



CCB. In the sphere 2ir X O A = circumference of a 

 great circle. 



.'. Area of portion of a sphere = height of portion 

 X circumference of a great circle. 



If we imagine a right cylinder to be described about a 

 sphere, its curved area = circumference of a great circle 

 X diameter of sphere = area of sphere. 



Hence, "area of sphere = area of circumscribing 

 cylinder." 



Exam. : How many square miles of sea are visible 

 from the top of a mast SO feet above the surface 1 



Let O be the centre of the earth ; draw O A. O 0., so 

 that AC is perpendicular to OC. ; draw 

 C N perpendicular to O A. Then, if B N 

 = x. O B = r. The area visible from A 

 will be the area of the part of the sphere 

 whose depth is B N 



i. e. , will = 2-jr r z. 



Now, let A'B=p. Then, by similar tri- 

 angles 



r + p : r : : r : r x. 



* 



rp 



and visible area = 7 

 2-n-rp very nearly. 



7 



Now 



7r = 3-14159. 

 r = 3958 miles 



p = 80 feet = 



80 



1 



'66 



.'. visible area = 2 X 314159 X 3958 X 

 = 378 nearly. 



III. THE MENST7RATIOK OF SOLIDS. 



(1). If two solid Angles are each contained by three Plane 

 Angles, that are emial each to each, then the inclination 

 of the Planes will be equal, each to each. 



Let O, o, be two solid angles contained by the plane 

 angles A O B, BOG, C O A, and aob, boc, coa, respec- 

 tively. Then the plane A O B is inclined to plane COA 



Fig. 30. 



at the same angle that aob is inclined to aoc. For, take 

 O A = oo, and let the plane C A B be perpendicular to 

 O A, and cab perpendicular to oo. Then angles CAO, 

 B A O, coo, 600, are right angles ; and CAB, cab, aro 

 the inclination of the planes in question. 



Now in triangles AOB, 006. we have OA = oa and 

 angles BOA, OAB = angles boa, oab, each to each, 

 .'. B = 06 and AB = ab (Euclid, I., 26) ; similarly in 

 triangles COA, coa, we have O C = oc and A C = ac. 



Then in triangles BOC, boc, we have the sides 

 BO, OC = sides 60, oc, each to each, the included 

 angle BOC = 6oc. .'. (Euclid, I., 4) the base CB = 

 base cb. 



Hence, finally, in triangle ABC, abc, we have the 

 sides B A, A C = the sides ba, ac each to each, and the 

 base BC = base be. /. (Euclid, I., 8) CAB = cab. 

 Q. E. D. 



COB. 1. Hence (Fig. 30), if the solid angle O be super- 

 imposed on the solid angle o, so that A O coincides with 



Fig. 31. 



VOL. i. 



O oa, and the plane AOB with plane 006, then, because 

 angle AOB = angle 006 the line OB coincides with ob; 

 and because inclination of plane COA to A B equals 

 that of coa to cob, the plane COA will coincide with coa, 

 and hence O C with oc. 



COB. 2. If A B C D E F and abcdef are two prisms, the 

 edges of which are equal each to each, and the plane 

 angles at B equal those at 6, the prisms are equal in all 

 respects. (See Figs. 31 and 32). 



For, if the triangle A B C be applied to abc so that 

 A B coincides with ab, and B with be, it is plain by 

 the last corollary that BD coincides with bd. And 

 hence the prisms will coincide throughout. 



4 o 



