...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...

...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...

...others adapted particularly to surveying, the description and use of which will be found in the notes. 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,...

...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...

...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,...

...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...

...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...

...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...

...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...

...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...