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. 1872 Excerpt: ...AE and AI. 1. Let the altitudes be commensurable, and suppose, for example, that AE is to AI, as 15 is to 8. Conceive AE to be divided into 15 equal parts, of which AI will contain 8; through the points of division let planes be passed parallel to ABCD. These planes will d1vide the parallelopipedon AG into 15 parallelopipedons, which have equal bases (P. II. C.) and equal altitudes; hence, they are equal (P. X., Cor. 3). Now, AG contains 15, and AL 8 of these equal parallelopipedons; hence, A G is to AL, as 15 is to 8, or as AE is to AI. In like manner, it may be shown that AG is to AL, as AE is to AI, when the altitudes are to each other as any other whole numbers. 2. Let the altitudes be incommensur-B C able. Now, if AG is not to AL, as AE is to AI, Lt t1s suppose that, AG: AL:: AE: AO, in which AO is greater than AI. Divide AE into equal parts, such that each shall be less than 01; there will be at least one po1nt of division m, between 0 and L Let P denote the parallelopipedon, whose base is ABCD, and altitude Am; since the altitudes AE, Am, are to each other as two whole numbers, we have, AG: P:: AE: Am. But, by hypothesis, we have,-4?: AL::: 40; therefore (B. II., P. IV., C), AL: P:: AO: Am. But AO is greater than Am; hence, if the proportion is true, AL must be greater than P. On the contrary, it is less; consequently, the fourth term of the proportion cannot be greater than AL In like manner, it may be shown that the fourth term cannot be less than AI; it is, therefore, equal to AL In this case, therefore, AG is to AL, as AE is to AL Ilence, in all cases, the given parallelopipedons are to each other as their altitudes; which was to be proved. Sch. Any two rectangular parallelopipedons whose bases are equal in all their parts, are to each other as t...
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