 The area of a rectangle is equal to the product of its base and altitude. Given R a rectangle with base b and altitude a. To prove R = a X b. Proof. Let U be the unit of surface. .R axb U' Then 1x1 But - is the area of R.  Elementary Algebra - Page 9by Frederick Howland Somerville - 1908 - 407 pagesFull view - About this book
 | George Albert Wentworth - Geometry, Solid - 1899 - 248 pages
...drawn, the tangent is the mean proportional between the whole secant and its external segment. 398. The area of a rectangle is equal to the product of its base by its altitude. 400. The area of a parallelogram is equal to the product of its base by its altitude.... | |
 | William James Milne - Geometry, Modern - 1899 - 258 pages
...long and 7" wide ? How may the amount of surface, or the area of any rectangle, be found? . Theorem. The area of a rectangle is equal to the product of its base by its altitude. Data : Any rectangle, as A, whose base is d and altitude e. To prove area of... | |
 | Webster Wells - Geometry - 1899 - 450 pages
...M a ff a 165 304. The dimensions of a rectangle are its base and altitude. PROP. III. THEOREM. 305. The area of a rectangle is equal to the product of its base and altitude. Note. In all propositions relating to areas, the unit of surface (§ 302) is understood... | |
 | Harvard University - Geometry - 1899 - 39 pages
...of two rectangles are to each other as the products of their bases and their altitudes. Corollary. The area of a rectangle is equal to the product of its base and its altitude. THEOREM IV. The area of a parallelogram is equal to the product of its base... | |
 | Metal-work - 1901 - 548 pages
...found in this manner; hence, we have the following general statement: xJIJ. Area of 11 Kectiuigle. — The area of a rectangle is equal to the product of its base and altitude. This is called a general statement, because it is true for all rectangles. This... | |
 | Arthur Schultze - 1901 - 260 pages
...its sides 20 inches. Find the ratio of the areas of the two rectangles. PROPOSITION III. THEOREM 339. The area of a rectangle is equal to the product of its base and altitude. R 1 u Hyp. R is a rectangle with base 6 and altitude a, To prove area of R = ax... | |
 | Alan Sanders - Geometry, Modern - 1901 - 260 pages
...the first rectangle is 20 sq. 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. AD Let ABCD be any rectangle. To Prove ABCD = ax 6. Proof. Let the square U, each... | |
 | Edward Brooks - Geometry, Modern - 1901 - 278 pages
...factor S, We have R : R' = axb : a' X b'. II. 11. Therefore, etc. 150 PROPOSITION IV. — THEOREM. The area of a rectangle is equal to the product of its base by its altitude. Given. — Let R be a rectangle whose ba.se is b and altitude a. To Prove. —... | |
 | Thomas Franklin Holgate - Geometry - 1901 - 462 pages
...other words, P contains the unit area bh times, or the measure of P is bh. THEOREM. The measure of the area of a rectangle is equal to the product of its base and its altitude. Or, more briefly, the area of a rectangle is equal to the product of its base... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1902 - 394 pages
...Find the ratio of the areas of the two rectangles. 166 AREAS OF POLYGONS PROPOSITION III. THEOREM 339. The area of a rectangle is equal to the product of its base and altitude. R 1 u Hyp. R is a rectangle with base b and altitude a. To prove area of R = ax... | |
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