 | George William Myers - Mathematics - 1909 - 394 pages
...altitude; that is, A = \ba. Find A, if (1) 6 = 12 ft., and a = 16 feet; (2) 6=8.2 rd., and a=7.78 rods. 4. The area of a parallelogram is equal to the product of the base by the altitude; that is, A =ba. Find A, if (1) 6 = 28, and a = 19; (2) 6 = 16.3, and « = 14.6.... | |
 | George William Myers - Mathematics - 1910 - 304 pages
...that the areas of parallelograms having equal altitudes, are to each other as their bases. 4. Prove that the area of a parallelogram is equal to the product of the base and altitude. (Use Exercise 2.) 5. Prove that the area of a triangle is one-half the product of... | |
 | Wilbur Fisk Nichols - 1910 - 236 pages
...brevity we usually say, to find the area of a parallelogram multiply the length by the width. 3. Since the area of a parallelogram is equal to the product of the two dimensions, it follows that if the area and one dimension are given the other may be found. 4. Let... | |
 | Charles H. Gleason - Arithmetic - 1910 - 536 pages
...by the altitude. 2. The area of a rectangle is equal to the product of the length by the breadth. 3. The area of a parallelogram is equal to the product of the base by the altitude. MENSURATION 4. The area of a trapezoid is equal to £ the sum of the two parallel... | |
 | Geometry, Plane - 1911 - 192 pages
...that OQ is a third proportional to OP and 0.4. Find the locus of Q as P moves along the line. 5. Prove that the area of a parallelogram is equal to the product of its base and altitude. Prove that the area of a triangle is the same as that of a second triangle,... | |
 | George Wentworth, David Eugene Smith - Arithmetic - 1911 - 396 pages
...the parallelogram will be changed to a rectangle of the same base and the same altitude. Therefore, the area of a parallelogram is equal to the product of the numbers expressing the base and altitude. The base and the altitude must be expressed in the same units.... | |
 | James Charles Byrnes, Julia Richman, John Storm Roberts - Arithmetic - 1911 - 328 pages
...with the old figure in area ? Therefore, how may you find the area of a parallelogram ? 222. RULE. The area of a parallelogram is equal to the product of the base by the altitude expressed in like units : DENOMINATE NUMBERS altitude is 3 ft., the area of the... | |
 | Glenn Moody Hobbs - 1912 - 100 pages
...consequently, 2 , . , 2(A DXaltitude) . _ , . , the sum 01 these areas =— = A UXaltitude. 2, Rule: (b) The area of a parallelogram is equal to the product of the base and altitude. This will be true of any parallelogram and hence true of the rectangle, Fig. 34,... | |
 | George Albert Wentworth, David Eugene Smith - Arithmetic - 1912 - 374 pages
...the parallelogram will be changed to a rectangle of the same base and the same altitude. Therefore, the area of a parallelogram is equal to the product of the base and altitude. Tor example, if the base of a parallelogram is 3 ft. 8 in. and the altitude is 3... | |
 | George Morris Philips, Robert Franklin Anderson - Arithmetic - 1913 - 520 pages
...product of the number of units in the base and the number of units in the altitude; therefore: 448. The area of a parallelogram is equal to the product of the number of units in the base and the number of units in the altitude. NOTE. The base and altitude must... | |
| |
I label the two new points e and /." FIG. 2 With the help of this figure he then proceeds to the usual proof of the theorem that the area of a parallelogram is equal to the product of the base by the altitude, establishing the equality of certain lines... 
