 | Daniel Adams - Arithmetic - 1849 - 142 pages
...of square inches in the trapezoid. Hence, To find the area of a trapezoid. RULE;. Multiply one half the sum of the parallel sides by the perpendicular distance between them. EXAMPLES FOR PRACTICE. 1. What are the square contents of a board 12 feet long, 16 inches wide at one... | |
 | Charles Davies - Geometry - 1850 - 238 pages
...it will reach a window 50 feet from the ground : required the breadth of the street. Ans. 77,8875ft. PROBLEM VI. To find the area of a trapezoid. RULE....Required the area of the trapezoid /" ABCD, having given / _ _ _ A. EB AB=321,51 feet, £0=214,24 feet, and CE=171,16 feet. Operation. We first find the sum... | |
 | John Radford Young - Measurement - 1850 - 294 pages
...are sides of the two ends of the frustum ; the area of each trapezoidal face is found by multiplying the sum of the parallel sides by the perpendicular distance between them — that is, by the slant height of the frustum — and diriding by 2 ; and as the faces are all equal,... | |
 | Daniel Adams - Measurement - 1850 - 144 pages
...number of square inches in the trapezoid. Hence, To find the area of a trapezoid. Multiply one half the sum of the parallel sides by the perpendicular distance between them. EXAMPLES FOR PRACTICE. 1. What are the square contents of a board 12 feet long, 16 inches wide at one... | |
 | Oliver Byrne - Engineering - 1851 - 310 pages
...required. To find the area of a trapezoid, or a quadrangle, two of ivJwse opposite sides are parallel. — Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. Required the area of the trapezoid ABCD, whose sides AB and DC are... | |
 | Charles Davies - Geometry - 1886 - 340 pages
...a window 50 feet from ihf ground : required the breadth of the street. Ans. 77.8875 ft. PROBLEM V1. To find the area of a trapezoid. RULE. Multiply the sum of the parallel sides by the perpend1cular distance between them, and then divide the product by two : the quotient will be the... | |
 | Thomas Kentish - Geometrical drawing - 1852 - 272 pages
...equal to the product of the length and breadth. The area of a trapezoid is found by multiplying half the sum of the parallel sides by the perpendicular distance between them. A triangle is half a rectangle, and therefore its area is half the product of the height and base.*... | |
 | Ezra S. Winslow - Business mathematics - 1853 - 264 pages
...whereby required the distance from B to A. 4 : 15 :: 22 = 83| rods. Ans. ' OF TRAPEZOIDS AND TRAPEZIUMS. To find the area of a trapezoid. RULE. — Multiply the sum of the two parallel sides by the perpendicular distance between them, and divide the product by 2 ; the quotient... | |
 | Andrew Duncan (Surveyor) - Surveying - 1854 - 156 pages
...the farm, which is plain from the figure. 15th. To find the area of a Trapezoid Rule, multiply half the sum of the parallel sides by the perpendicular distance between them, and the product is the area. Let figure 12 be a Trapazoid ; if AD be a \-r bisected in E, and E " F drawn... | |
 | Thomas Kentish - Mathematical instruments - 1854 - 268 pages
...equal to the product of the length and breadth. The area of a trapezoid is found by multiplying half the sum of the parallel sides by the perpendicular distance between them. A triangle is half a rectangle, and therefore its area is half the product of the height and base.*... | |
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To find the area of a trapezoid. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and then divide the product by two : the quotient will be the area (Bk. 
