 | Andrew Wheeler Phillips, Irving Fisher - 1897 - 374 sider
...origin of the custom of calling the second power of a number its "square." PROPOSITION V. THEOREM 306. The area of a parallelogram is equal to the product of its base and alt1tude. GIVEN — the parallelogram ABCD, with base b and altitude a. To PROVE the area of ABCD =... | |
 | Middlesex Alfred Bailey - 1897 - 332 sider
...273, and 274. III. The area of a rectangle is equal to the product of its base by its altitude. IV. The area of a parallelogram is equal to the product of its base by its altitude. V. The area of a triangle is equal to one half the product of its base by its... | |
 | Silas Ellsworth Coleman - 1897 - 180 sider
...opposite side. A = 6a The altitude of a rectangle is equal to the side not taken as base. The area (A) of a parallelogram is equal to the product of its base and its altitude. EXPLANATION. This is a familiar fact in the case of rectangles ; the common form of statement... | |
 | James Howard Gore - 1898 - 232 sider
...linear units in its base by the number of linear units in its altitude. PROPOSITION IV. THEOREM. 251. The area of a parallelogram is equal to the product...its base and altitude. Let ABCD be a parallelogram. To prove that the area of £-2 ABCD = ABx AF. \ / Erect the perpendiculars AFsmd BE \l and produce... | |
 | Webster Wells - 1898 - 250 sider
...the product of 6 and 5, the numbers which express the lengths of the sides. PROP. IV. THKOREM. 309. The area of a parallelogram is equal to the product of its base and altitude. EB FC A h D Given O ABCD, with its altitude DF = a, and its base AD=b. To Prove area ABCD = a x b.... | |
 | John Henry Tanner, Joseph Allen - 1898 - 458 sider
...about the ellipse ; its sides are parallel to, and equal in length to, the conjugate diameters. Since the area of a parallelogram is equal to the product of its adjacent sides and the sine of the included angle, therefore the area of this circumscribed parallelogram... | |
 | Webster Wells - 1899 - 450 sider
...the product of 6 and 5, the numbers which express the lengths of the sides. PROP. IV. THEOREM. 309. The area of a parallelogram is equal to the product of its base and altitude. EBFC AI) D Given O ABCD, with its altitude DF= a, and its base AD = b. To Prove area ABCD = ax b. Proof.... | |
 | Charles Austin Hobbs - 1899 - 266 sider
...unit of surface ; that is, the area is equal to 6 x 4 units of surface. Proposition 165. Theorem. 201. The area of a parallelogram is equal to the product of its base and altitude. A a Hypothesis. ABCD is a O whose base and altitude are AB and BE respectively. Conclusion. Area of... | |
 | Webster Wells - 1899 - 424 sider
...the product of 6 and 5, the numbers which express the lengths of the sides. PROP. IV. THEOREM. 309. The area of a parallelogram is equal to the product of its base and altitude. BFC A b D Given O ABCD, with its altitude DF=a, and its base To Prove area ABCD = ax b. Proof. Draw... | |
 | William Taylor Campbell - 1899 - 276 sider
...feet wide instead of 27 feet. How much additional area does it contain f 2. Area of a Parallelogram. The area of a parallelogram is equal to the product of its base and altitude. The base is any one of the sides. The altitude is the perpendicular distance between the base and the... | |
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The area of a parallelogram is equal to the product of its base and altitude. 
