| Mathematics - 1801 - 658 pages
...street. • t Ans. 76- 1 2333 35 feet. PROBLEM IV. 7o f:nd tlie area of a trapezoid. • RULE.* Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. • EXAMPLES. * DEMONSTRATION. or (because B»=DE) =-, .-. A ABD+... | |
| Abel Flint - Surveying - 1804 - 226 pages
...8925X0.47076=4201 the double Area of the Triangle. PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the Sum of the two parallel Sides by half the perpendicular distance ; the Product will be the Area.... | |
| Abel Flint - Surveying - 1808 - 190 pages
...the double Area of the Triangle. • PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the sum of the two parallel Sides by half the perpendicular distance ; the Product will be the Area.... | |
| Thomas Keith - 1817 - 306 pages
...acres. <• 2 roods 21 perches. PROBLEM VIII. • To find the Area of a Trapezoid. RULE *. Multiply half the sum of the two parallel sides by the perpendicular...distance between them, and the product will be the area. Example 1. Let AB c D JE. be a trapezoid, the side '-. A )•. — 23, D c = 9•5, and CI — 13,... | |
| Anthony Nesbit, W. Little - Measurement - 1822 - 916 pages
...Ant. 97.3383 bushels. PROBLEM VII. To Jind the area of a trapezoid. RULE. • By the Pen. Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area in square inches. Divide this area by 2 82, 231, and 2150.42, and... | |
| John Nicholson - Machinery - 1825 - 822 pages
...square 63 I 189 of AB has been subtracted. 3 I 189 Prob. 4. To find the Area of aTrapezoid. Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. Ex. In a trapezoid, the parallel sides are AB 7, and CD 12, and... | |
| Abel Flint - Surveying - 1825 - 252 pages
...0.47076=4201 tbe double Area of the Triangle. PROBLEM X. To find the Jbeaof a TrapezoiA. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the sum of the two parallel Sides by half the perpendicular distance, the product will be the Area.... | |
| Zadock Thompson - Arithmetic - 1826 - 176 pages
...the area ? Ans. 54.299 rods. I Problem III. Tojind the area of a trapezoid. :BuLE.— Multiply half the sum of the two parallel sides by the perpendicular...distance between them, and the product will be the area. Examples. 1. One of the two parallel sides of a trapezoid is 7.5 chains and the other 12.25, and the... | |
| John Nicholson (Civil engineer) - Building - 1830 - 240 pages
...square 63 I 189 of AB has been subtracted. 3 I 189 Prob. 4. To find the Area of aTrapezoid. Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. Ex. In a trapezoid, the parallel sides are AB 7, and CD 12, and... | |
| Ira Wanzer - Arithmetic - 1831 - 408 pages
...40, and 50 rods? Ans. 3f A. PROBLEM V, — To find the area of a Trapezoid. RULE. — Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the pr«' duct will be the area. Ex. How many square feet are contained in a board which is 12 feet 6 inches... | |
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