 | 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. —... | |
 | Frank Joseph Schneck - Business mathematics - 1902 - 288 pages
...and three or more sides that are parallelograms, is a Prism. TRIANGULAR PRISM RECTANGULAR PRISM 212. The area of a rectangle is equal to the product of its length and breadth. 213. A rectangular prism that is one unit high has a volume equal to the product... | |
 | American School (Chicago, Ill.) - Engineering - 1903 - 390 pages
...that is, B ~ tf Multiplying equations (1) and (2), we have A a X b B ~ a' x V THEOREM 1 XIII. 1 94. The area of a rectangle is equal to the product of its base and altitude. iL Let a and b be the numerical measures of the altitude and base of the rectangle A, and let B be... | |
 | Alan Sanders - Geometry - 1903 - 396 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. BC D Let ABCD be any rectangle. To Prove ABCD = axb. Proof. Let the square U, each side of which is... | |
 | Clara Avis Hart, Daniel D. Feldman - Geometry - 1912 - 504 pages
...of the Greeks, which was logical and deductive, even from its beginning. PROPOSITION I. THEOREM 475. The area of a rectangle is equal to the product of its base and its altitude. (See § 476.) BC V AD « Given rectangle ABCD, with base AD and altitude AB, and let... | |
 | William Benjamin Fite - Algebra - 1913 - 304 pages
...mW + 5 mw4 + w5. 51. Multiplication of Polynomials. — The student is familiar with the fact that the area of a rectangle is equal to the product of its base and altitude. If we have two rectangles with the common altitude a and bases x and y respectively, their combined... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry, Plane - 1913 - 328 pages
...of its sides 20 in. Find the ratio of the areas of the two rectangles. PROPOSITION III. THEOREM 347. 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.... | |
 | William Benjamin Fite - Algebra - 1913 - 368 pages
...m2»3 + 5 mn* + и5. 51. Multiplication of Polynomials. — The student is familiar with the fact that the area of a rectangle is equal to the product of its base and altitude. If we have two rectangles with the common altitude a and bases x and y respectively, their combined... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 490 pages
...of its sides 20 in. Find the ratio of the areas of the two rectangles. PROPOSITION III. THEOREM 347. 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 B = a x b. Proof. Let U be the unit of surface.... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Solid - 1913 - 184 pages
...called its dimensions. In Chapter IV (§ 181), we assumed (without proof) the well-known principle that the area of a rectangle is equal to the product of its two dimensions. Similarly, we shall now assume that the volume of a rectangular parallelepiped" is... | |
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The area of a rectangle is equal to the product of its base and altitude. 
