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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1827 Excerpt: ... ABCD and vertex the point L, shall have to the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG. For if the cone ABCDL has not to the cone EFGHN the triplicate ratio of that which AC has to EG, the cone ABCDL must have the triplicate of thnt ratio to some solid which is less or greater than the cone EFGHN. First, let it have it to a less, viz. to the solid X. Make the same construction as in the preceding proposition, and it may be demonstrated the very same way as in that proposition, that the pyramid of which the base is the polygon EOFPGRHS, and vertex N, is greater than the solid X. Inscribe also in the circle ABGD the polygon ATBYCVDQ similar to the polygon EOFPGRHS, upon which erect a pyramid having the same vertex with the cone; and let LAQ be one of the tanglesj containing the pyramid upon the polygon A1BYCVLQ, the vertex of which is L; and let NES be one of the triangles containing the pyramid upon the polygon EOFPGRHS of which the vertex is N; and join KQ, MS. Then, because the cone ABCDL is similar to the cone EFGHN, AC is to EG as the axis KL to the axis MN; and as AC to EG, so is AK to EM; therefore as AK to EM, so is KL to MN; and alternately, AK to KL, as EM to MN: and the right angles AKL, EMN, are equal: therefore, the sides about these equal angles being proportionals, the triangle AKL is similar to the triangle EMN. Again, because AK is to KQ, as EM to MS, and that these sides are about equal angles AKQ, EMS, because these angles are, each of them, the same part of four right angles at the centres K, M; therefore the triangle AKQ is similar to the triangle EMS. And because it has been shewn that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS: therefore as QK to...
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