Articoli correlati a Elements of geometry and trigonometry; from the works...
Elements of geometry and trigonometry; from the works of A. M. Legendre. Adapted to the course of mathematical instruction in the United States - Brossura
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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. 1864 Excerpt: ...oblique parallelopipedon is equal in volume to rectangular parallelopipedon, having an equal base and an qual altitude. Cor. 3. Any two parallelopipedons are equal in volume, when they have equal bases and equal altitudes. PROPOSITION XI. THEOREM. Two rectangular parallelopipedons having a common lower base, are to each other as their altitudes. Let the parallelopipedons AG and AL have the common lower base AB CD: then will they be to each other as their altitudes 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 AB CD. These planes will divide 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 g jj 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. m 2. Let the altitudes be incommensurable. K B Now, if AG is not to AL, as AE is to AI, let us suppose that, AG: AL.:: AE: AO, in which AO is greater than AI. Divide AE into equal parts, such that each shall b« ess than 01; there will be at least one point of division m, between 0 and I. Let P denote the parallelopipedon, whose base is AB CD, and altitude Am; since the altitudes AE, Am, are to each other as two whole numbers, we have, K AG: P:: AE: Am. But, by hypothesis, we have, AG: AL:: AE: AO; therefore (B. H, P. IV., C.), AL: P:: AO: Am. B:C But AO is greater than Am; hence, if the proportion is true, AL must be greater...
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