 | Frank Joseph Schneck - Business mathematics - 1902 - 288 pages
...the rectangle? How does the altitude of one ( , j. compare with that of the other? Principle.—The area of a parallelogram is equal to the product of its base and altitude. PROBLEMS 1. Find the area of a parallelogram whose base is 16 rd. and whose altitude is 8 rd. 2. What... | |
 | George Albert Wentworth - Geometry, Solid - 1902 - 248 pages
...drawn, the tangent is the mean proportional between the whole secant and its external segment. 400. The area of a parallelogram is equal to the product of its base by its altitude. 401. Parallelograms having equal bases and equal altitudes are equivalent. 403.... | |
 | American School (Chicago, Ill.) - Engineering - 1903 - 390 pages
...195. Corollary. The area of a square is equal to the square of one of its sides. THEOREn LXIV. Ip6. The area of a parallelogram is equal to the product of its base and altitude. Let ABCD be a parallelogram leaving its altitude DF equal to a and its base AD equal to ft. To prove that ABCD... | |
 | Alan Sanders - Geometry - 1903 - 392 pages
...ft., as that has not yet been established.] PROPOSITION V. THEOREM 586. The area of a rectangle is equal to the product of its base and altitude. Let ABCD be any rectangle. To Prove ABCD = ax 6. Proof. Let the square U, each side of which is a linear ui.it, be... | |
 | Alan Sanders - Geometry - 1903 - 396 pages
...in the rectangle is 4 x 3 or 12. The area then is 12 square feet. 588. COROLLARY I. The area of any parallelogram is equal to the product of its base and altitude. Let A BCD be any parallelogram BE C and DE be its altitude. r To Prove ABCD = AD x DE. Proof. Draw AF _L... | |
 | George E. Mercer, Mabel Bonsall - Arithmetic - 1914 - 324 pages
...should be given practice in drawing them until the mention of a name calls forth an appropriate drawing. The area of a parallelogram is equal to the product of its base and altitude. The base and altitude must be expressed in units of the same kind. The diagonal of a parallelogram... | |
 | Claude Irwin Palmer, Daniel Pomeroy Taylor - Geometry, Plane - 1915 - 320 pages
...rt.AAED = rt.ABFC. Why? ABFD - AAED = nABFE. ABFD - ABFC = OABCD. • .'. nABFE = UABCD. Why? 355. Theorem. The area of a parallelogram is equal to the product of its base and altitude. As a formula, A = bh. This follows from §§ 346 and 354. 356. Theorem. The altitude of a parallelogram... | |
 | John Wesley Young, Albert John Schwartz - Geometry, Modern - 1915 - 248 pages
...altitude is then the perpendicular distance between the base and the opposite side (§§ 228, 229). 348. THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. E n FIG. 162. Given the parallelogram ABCD with the base 6 and the altitude h.... | |
 | Edward Rutledge Robbins - Geometry, Plane - 1915 - 282 pages
...times as long as it is wide, how many square yards will the field contain? PROPOSITION IV. THEOREM 359. The area of a parallelogram is equal to the product of its base by its altitude. FD EC Given : O ABCD, with base b and j altitude h. To Prove : Area of ABCD =... | |
 | William Betz - Geometry - 1916 - 536 pages
...144 250 300 t 8 6 10 10 I II III IV P 12 16 30 2a+ 26 A 8 15 66 06 AREAS PROPOSITION I. THEOREM 328. The area of a parallelogram is equal to the product of its base and altitude. EC Given the parallelogram ABCD, with its base AB equal to ft, and its altitude BE equal to h. To prove... | |
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The area of a parallelogram is equal to the product of its base and altitude. 
