| John Farrar - 1822 - 270 pages
...the product of any two contiguous sides. 64. if the parallelogram be oblique.angled, as ABCD (Jig. 64), the breadth or perpendicular distance of either...radius being unity. Given AB = 59 chains 80 links, or 59,80ch., AC = 37,05 ch., and A = 90°, we have 59,8 x 37,05 x 1 = 2215,59 square chains = 22155900... | |
| John Farrar - 1822 - 244 pages
...product of any two contiguous sides. Fig. 64. If the parallelogram be oblique-angled, as ABCD (Jig. 64), the breadth or perpendicular distance of either...radius being unity. Given AB = 59 chains 80 links, or 59,80ch., AC = 37,05 ch., and A — 90°, we have 59,8 x 37,05 X 1 = 2215,59 square chains = 2:2155900... | |
| John Farrar - 1833 - 274 pages
...supplements of each other (Geom. 64). The sines of all the angles therefore are equal to each other. product of the length by the breadth is the product...= 37,05 ch., and A = 90°, we have 59,8 X 37,05 X 1 = 2215,59 square chains = 22155900 square links. Now, since 10 square chains, 100000 square links,... | |
| John Farrar - 1840 - 270 pages
...product of any two contiguous sides. Fig. 64. If the parallelogram be oblique-angled, as AB CD (Jig. 64), the breadth or perpendicular distance of either...= 37,05 ch., and A = 90°, we have 59,8 X 37,05 X 1 = 2215,59 square chains = 22155900 square links. Now, since 10 square chains, 100000 square links,... | |
| Nicholas Tillinghast - 1844 - 110 pages
...Every parallelogram is equivalent to a rectangle which has an equal base and equal altitude. Cor. 2. Hence the area of a parallelogram is equal to the product of its base by its altitude (Prop. Cor. 3. Hence parallelograms of equal altitudes, are m proportion to... | |
| Thomas Lund - 1854 - 520 pages
...measure of the parallelogram ABCD=ABxBE = the base*. the height, as it is usually stated. In other words the area of a parallelogram is equal to the product of any one side and the perpendicular distance of that side from the opposite side. In any proposed case,... | |
| John Radford Young - 1855 - 218 pages
...parallelogram, and CE its altitude ; then, in the right-angled triangle AEC, we shall have _^* BO that the area of a parallelogram is equal to the product of any two adjacent sides multiplied by the sine of the angle between them. The trouble of finding the perpendicular... | |
| Henry Bartlett Maglathlin - 1881 - 418 pages
...diagram that the rhombus ABCD is equal to the rectangle EBCF of the same base and altitude (Art. 218). Hence, The area of a parallelogram is equal to the product of the base and altitude. 9. What is the area of a parallelogram whose base is 36 feet and altitude 15... | |
| 1883 - 248 pages
...ABC = -ft and (1.) p = b sin A, which substitute for p, be sin A ac sin B ab sin C area of ABO = 2 Hence, the area of a parallelogram is equal to the product of two adjacent sides into the sine of the angle between them. Also, since the sine of an angle is equal... | |
| Joseph Johnston Hardy - 1897 - 398 pages
...cos A sin B. 14. The sides of a triangle are proportional to the sines of the opposite angles. 15. The area of a parallelogram is equal to the product of any two adjacent sides by the sine of the included angle. 16. In quadrant 2 the sine and cosecant only are... | |
| |