 | Benjamin Greenleaf - Geometry - 1873 - 202 pages
...rectangles ABCD, AE FD, having equal altitudes, are to each other as their bases AB, AE. THEOREM IV. 185. Any two rectangles are to each other as the products of their bases multiplied by their altitudes. Let ABCD, AEGF be two DC rectangles ; then will ABCD be toAEGFusAB multiplied by AD is to AE multiplied... | |
 | Aaron Schuyler - Measurement - 1864 - 506 pages
...incommensurable, denote the area by k', the base by 6', and the altitude by a'. Then, since by Geometry any two rectangles are to each other as the products of their bases and altitudes, we have k : k' :: ab : a'b'. But k — ah, .-. k' = a'b'. 159. Problem. To find the... | |
 | Benjamin Greenleaf - Geometry - 1874 - 206 pages
...rectangles AB CD, A EFD, having equal altitudes, are to each other as their bases A B. AE. THEOREM IV. 185. Any two rectangles are to each other as the products of their bases multiplied by their altitudes. Let AB CD, AEGF be two 5- ° ° rectangles ; then will ABCD be i to AEGF as AB multiplied by j , ,... | |
 | Euclides - 1874 - 342 pages
...parallelogram CF; ex sequali, 4. K is to M, as the parallelogram AC to the parallelogram CF(V. 22); but .5" has to M the ratio which is compounded of the ratios of the sides : therefore also 5. The parallelogram AC has to the parallelogram CF, the ratio which is compounded of the ratios of... | |
 | Euclid, James Bryce, David Munn (F.R.S.E.) - Geometry - 1874 - 236 pages
...together, we have — ABO _ AB AC ADE " AD x AE' ABC—AB AC. OPQ ~ OP * OQ' that is, the triangles are in the ratio which is compounded of the ratios of the sides> Cor. 1. — If the triangles are similar, that is, similar triangles are as the squares on their homologous... | |
 | Aaron Schuyler - Measurement - 1875 - 284 pages
...incommensurable, denote the area by k', the base by b', and the altitude by a'. Then, since by Geometry any two rectangles are to each other as the products of their bases and altitudes, we have k : k' : : ab : a'b'. But k = ab, .-. k' = a'b'. 159. Problem. To find tJie... | |
 | 1875 - 256 pages
...Proof in both cases. 2. To make a square which is to a given square in a given ratio. 3. Prove that two rectangles are to each other as the products of their bases by their altitudes. What follows if we suppose one of the rectangles to be the unit of surface ? 4.... | |
 | Robert Potts - Geometry - 1876 - 446 pages
...parallelogram CF; ex fequali, K is to M, as the parallelogram ACio the parallelogram CF: (v. 22.) but K has to M the ratio which is compounded of the ratios of...ratio which is compounded of the ratios of the sides Wherefore, equiangular parallelograms, &c. QED PROPOSITION XXIV. THEOREM. Parallelograms about the... | |
 | Euclid - 1876 - 240 pages
...parallelogram CF ; ex cequali, K is to M, as the parallelogram AC is to the parallelogram CF (/). But K has to M the ratio which is compounded of the ratios of the sides ; therefore also the parattelogram AC has to .the parallelogram CF the ratio which is compounded of the ratios of the sides.... | |
 | William Guy Peck - Conic sections - 1876 - 412 pages
...in its base multiplied by the number of linear units in its altitude, which was to be proved. Cor. Any two rectangles are to each other as the products of their bases and altitudes ; if their bases are equal, they are to each other as their altitudes. Scho. The product... | |
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K hag to M the ratio which is compounded of the ratios of the sides ; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. COR. Hence, any two rectangles are to each other as the products... 
