 ... 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.  Plane and Solid Geometry - Page 155by Claude Irwin Palmer, Daniel Pomeroy Taylor - 1918 - 436 pagesFull view - About this book
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 490 pages
...QBD 351. COR. 1. Parallelograms having equal bases and equal altitudes are equivalent. 352. COR. 2. Any two parallelograms are to each other as the products of their bases and altitudes. 353. COR. 3. Parallelograms having equal bases are to each other as their altitudes. 354.... | |
 | Walter Burton Ford, Earle Raymond Hedrick - Geometry, Modern - 1913 - 272 pages
...O ABCD = rectangle ABGF. But rectangle ABGF=ab, §181 whence, O ABCD = ab. 1/187. Corollary 1. (a) Two parallelograms are to each other as the products of their bases and altitudes. (6) Two parallellograms that have equal bases and equal altitudes are equal in area. PARALLELOGRAMS... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Solid - 1913 - 184 pages
...area of a parallelogram is equal to the product of its base by its altitude. 187. Corollary 1. (a) Two parallelograms are to each other as the products of their bases and altitudes. 188. Corollary 2. Two parallelograms that have equal altitudes are to each other as their... | |
 | George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 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. PROPOSITION V. THEOREM 325. The area of a triangle is equal to half the product... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Plane - 1913 - 376 pages
...ABCD = rectangle ABGF. But rectangle ABGF= ab, § 181 whence, O ABCD — ab. 187. Corollary 1. (a) Two parallelograms are to each other as the products of their bases and altitudes. (b) Two parallellograms that have equal bases and equal altitudes are equal in area. 164... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Plane - 1913 - 378 pages
...ABCD = rectangle ABGF. But rectangle ABGF= ab, § 181 whence, O ABCD = ab. 187. Corollary 1. (a) TVo parallelograms are to each other as the products of their bases and altitudes. (b) Two parallellograms that have equal bases and equal altitudes are equal in area. EXERCISES... | |
 | Claude Irwin Palmer, Daniel Pomeroy Taylor - Geometry, Plane - 1915 - 336 pages
...Theorem. The base of a parallelogram is equal to its area divided by its altitude. As a formula, 6 = 1-. h 358. Theorem. Parallelograms having equal bases and...material as a yard cut at right angles to the edge? Why? AGB FIG. 3 3. Through a point in a diagonal of a parallelogram, lines are drawn parallel to the sides,... | |
 | Claude Irwin Palmer, Daniel Pomeroy Taylor - Geometry, Plane - 1915 - 296 pages
...triangles having equal altitudes are to each other as their bases. 369. Theorem. Any two triangles are to each other as the products of their bases and their altitudes. EXERCISES 1. In Fig. 1, AB II CD. Prove that AEH _K_ A CED, CHD, and CKD are equivalent. \i 2. Prove that a median... | |
 | Claude Irwin Palmer, Daniel Pomeroy Taylor - Geometry, Plane - 1915 - 320 pages
...triangles having equal altitudes are to each other as their bases. 369. Theorem. Any two triangles are to each other as the products of their bases and their altitudes. Fio. 1 EXERCISES 1. In Fig. 1, AB II CD. Prove that AE A CED, CHD, and CKD are equivalent. ~^ 2. Prove... | |
 | Webster Wells, Walter Wilson Hart - Geometry, Plane - 1915 - 330 pages
...base 62 and altitude a.2. (1) Parallelograms having equal bases and equal altitudes are equal. (2) Two parallelograms are to each other as the products of their bases by their altitudes. For, since O Pi = atbi and a P2 = o262, then ^^ = ^-l . O P.2 O2&2 (3) Parallelograms... | |
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