 | Mathematics - 1801 - 658 pages
...the 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...between them, and half the product will be the area. • EXAMPLES. * DEMONSTRATION. or (because B»=DE) =-, .-. A ABD+ A BCD, or , , , ABxDE , DCxDE the... | |
 | 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.... | |
 | James Thompson - Arithmetic - 1808 - 176 pages
...trafiezoid, or quadrangle, <u'o cf •whose opposite sides are parallel. RULE — Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product •will be the area. EXAMPLES. 13. Required the area of a trapezoid whose parallel sidef »re 25 feel 6 inches, and 18 feet... | |
 | Samuel Webber - Mathematics - 1808 - 466 pages
...breadth of the street. Ans. 76' 1233335 feet. 414PR0BLEM IV. To find the area of a trapezold. RULE.* Multiply the sum of the two parallel sides by the perpendicular distance between them, and hall" the product will be the area. EXAMPLES. 1. In a trapezoid, the parallel sides are AB 7' 5, and... | |
 | 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,... | |
 | Anthony Nesbit, W. Little - Measurement - 1822 - 916 pages
...the grain? 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...between them, and half the product will be the area in square inches. Divide this area by 2 82, 231, and 2150.42, and the respective quotients will be... | |
 | Anthony Nesbit - Surveying - 1824 - 476 pages
...Ans. 1131^.2 in. 9 pa. PROBLEM VIII. To find the area of a Irapezoid. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. Or, half the sum of the sides multiplied by their distance will give the area. EXAMPLE. 1. What is... | |
 | 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... | |
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Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. 
