 | John Farrar - Logarithms - 1822 - 270 pages
...equal to each other. product of the length by the breadth is the product of any two contiguous sides. 64. if the parallelogram be oblique.angled, as ABCD...= 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, or 100000 square links,... | |
 | John Farrar - Trigonometry - 1833 - 276 pages
...product of any two contiguous sides. Fig. 64. If the parallelogram be oblique-angled, as ABCD (fig- 64), the breadth or perpendicular distance of either...radius being unity. Given AB = 59 chains 80 links, or 59,80 ch., AC = 37,05 ch., and A = 90°, we have 59,8 X 37,05 X 1 = 2215,59 square chains = 22155900... | |
 | John Farrar - Trigonometry - 1833 - 274 pages
...product of any two contiguous sides. Fig. 64. If the parallelogram be oblique-angled, as ABCD (fig- 64), the breadth or perpendicular distance of either...radius being unity. Given AB = 59 chains 80 links, or 59,80 ch., AC = 37,05 ch., and A = 90°, we have 59,8 X 37,05 X 1 = 2215,59 square chains = 22155900... | |
 | John Farrar - Logarithms - 1840 - 270 pages
...the 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...radius being unity. > Given AB = 59 chains 80 links, or 59,80 ch., AC = 37,05 ch., and A = 90°, we have 59,8 X 37,05 X 1 = 2215,59 square chains = 22155900... | |
 | Nicholas Tillinghast - Geometry, Plane - 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 - Geometry - 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 - Arithmetic - 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 - Geometry, Analytic - 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... | |
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ABCD (fig- 64), the breadth or perpendicular distance of either two opposite sides, as CP, is equal to the product of the corresponding oblique side CB by the sine of the angle of the parallelogram, radius being unity (Trig. 30). Hence, the area of a... 