 ... any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 403. The area of a triangle is equal to half the product of its base by its altitude.  Elements of Plane and Solid Geometry - Page 179by George Albert Wentworth - 1877 - 398 pagesFull view - About this book
 | George Albert Wentworth - Geometry, Modern - 1881 - 266 pages
...and C' 0'. Then АACB = ¿ВXС0 =^_x_CO_, § 326 Л A' С' B' A' B' X C' 0' A' B' C" O' (two A are to each other as the products of their bases by their altitudes). But -- = , § 297 A'B' С'O' (the homologous altitudes of similar A have the same ratio as their homologous... | |
 | George Albert Wentworth - Geometry, Modern - 1882 - 268 pages
...having equal bases and equal altitudes are equivalent. 323. Cor. 2. Parallelograms having equal hases are to each other as their altitudes ; parallelograms...a triangle is equal to one-half of the product of Us base li'/ its altitude. Let ABC be a triangle, AB its base, and CD its altitude. We are to prove... | |
 | Alfred Hix Welsh - Geometry - 1883 - 326 pages
...one-half of any parallelogram having an equal base and an equal altitude. Cor. II.—Any two triangles are to each other as the products of their bases by their altitudes. For, let T and T' denote two triangles whose bases are b and b', and whose altitudes are a and a'.... | |
 | Webster Wells - Geometry - 1886 - 392 pages
...prove that -A^- = -^MA'B'C' A'B'2 Draw the altitudes CD and C'D'. Then since any two triangles are to each other as the products of their bases by their altitudes (§ 329), we have ABC AB x CD AB CD x A'B'C' A'B' x C'D' A'B' C'D' But the homologous altitudes of... | |
 | William Chauvenet, William Elwood Byerly - Geometry - 1887 - 331 pages
...to be understood " surface of the rectangle." PROPOSITION III.—THEOREM. 7. Any two rectangles are to each other as the products of their bases by their altitudes, Let E and R' be two rectangles, k and k their bases, h and h ' their altitudes; then E _ k XA R ' Jfx... | |
 | George Albert Wentworth - Geometry - 1888 - 272 pages
...other as their altitudes ; parallelograms having equal altitudes are to each other as their bases ; any two parallelograms are to each other as the products of their bases by their altitudes. AREAS OF POLYGONS. PROPOSITION V. THEOREM. 368. The area of a triangle is equal to one-half of the... | |
 | George Albert Wentworth - Geometry - 1888 - 264 pages
...the altitudes, AD and AD as the bases. PROPOSITION II. THEOREM. 362. The areas of two rectangles are to each other as the products of their bases by their altitudes. r s L_ b V b i Let R and R' be two rectangles, having for their bases b and b', and for their altitudes... | |
 | Edward Albert Bowser - Geometry - 1890 - 420 pages
...equal altitudes are equivalent, because they are all equivalent to the same rectangle. 365. COR. 2. Any two parallelograms are to each other as the products of their bases by their altitudes; therefore parallelograms having equal bases are to each other as their altitudes, and parallelograms... | |
 | George Albert Wentworth - Geometry - 1892 - 468 pages
...altitudes; parallelograms having equal altitudes are to each other as their bases; any two parallelograms an to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 368. The area of a triangle is equal to one-half of the product of its base by its altitude. H I B... | |
 | Nicholas Murray Butler, Frank Pierrepont Graves, William McAndrew - Education - 1892 - 544 pages
...criticism, it will be necessary to reproduce the demonstration given. To prove that two rectangles are to each other as the products of their bases by their altitudes. R and R' are two rectangles, having for their bases b and b' and for their altitudes a and a'. It is... | |
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