 | American School (Chicago, Ill.) - 1903 - 390 σελίδες
...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... | |
 | Fletcher Durell - 1904 - 382 σελίδες
...area of a rectangle 8 yds. long and 5 ft. wide. BOOK IV. PLANE GEOMETRY PROPOSITION IV. THEOREM 385. The area of a parallelogram is equal to the product of its base by its altitude. KB FC Given the ZZ7 ABCD with the base AD (denoted by 6) and the altitude DF (denoted by... | |
 | Fletcher Durell - 1904 - 232 σελίδες
...side. BOOK IV. 383. The area of a rectangle is equal to the product of its base by its altitude. 385. The area of a parallelogram is equal to the product of its base by its altitude. 389. The area of a triangle is equal to one-half the product of its base by its altitude.... | |
 | Fletcher Durell - 1911 - 553 σελίδες
...Find, in square feet, the area of a rectangle 8 yds. long and 5 ft. wide. PROPOSITION IV. THEOREM 385. The area of a parallelogram is. equal to the product of its base by its altitude. K__B__ pa A b D Given the ZZ7 ABCD with the base AD (denoted by b) and the altitude DF (denoted... | |
 | George Albert Wentworth - 1904 - 496 σελίδες
...evident by dividing the figure into squares, each BOOK IV. PLANE GEOMETRY. PROPOSITION IV. THEOREM. 400. The area of a parallelogram is equal to the product of its base by its altitude. CF CE A b DA b D Let AEFD be a parallelogram, b its base, and a its altitude. To prove that... | |
 | George Clinton Shutts - 1905 - 260 σελίδες
...fractional number of times in the base and altitude the same rule is reached. PROPOSITION IV. 330. Theorem. The area of a parallelogram is equal to the product of its base and altitude. Suggestion. § 323 and § 326. PROPOSITION V. 331. Theorem. The area of a triangle... | |
 | Elmer Adelbert Lyman - 1905 - 268 σελίδες
...C'fi, thus forming a rectangle with the, same base and altitude as the given parallelogram. Therefore,^ the area of a parallelogram is equal to the product of its base and altitude, or A = ab. 188. The Triangle. Since the line AB divides the parallelogram into two... | |
 | Edward Rutledge Robbins - 1906 - 268 σελίδες
...QED 372. THEOREM. The area of a square is equal to the square of its side. (See 371.) 373. THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. Given : L3 ABCD whose base is b and altitude, h. To Prove : Area of ABCD = b - h. Proof:... | |
 | William James Milne - 1906 - 364 σελίδες
...550. The area of any parallelogram is equal to that of a rectangle having the same base and altitude. Hence, The area of a parallelogram is equal to the product of its base and altitude. 551. Find the area of each of these parallelograms : BASE ALTITUDE BASE ALTITUDE... | |
 | William James Milne - 1906 - 364 σελίδες
...550. The area of any parallelogram is equal to that of a rectangle having the same base and altitude. Hence, The area of a parallelogram is equal to the product of its base and altitude. 551. Find the area of each of these parallelograms : BASE ALTITUDE BASE ALTITUDE... | |
| |
The area of a parallelogram is equal to the product of its base and its height: A = bx h. 
