| 1801 - 426 sider
...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 - 1804 - 168 sider
...and 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 - 1808 - 168 sider
...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.... | |
| Peter Nicholson - 1809
...42 504=the area of ABCD. PROBLEM VI. To find the area of a trapezoid. Multiply the half sum of the **parallel sides by the perpendicular distance between them, and the product will be the area.** EXAMPLE I. What is fhe area of a board or plank in the form of a trapeziod, being 1f. 7i. one end,... | |
| Thomas Keith - 1817
...17•6327 acres; or 17 acres. <• 2 roods 21 perches. PROBLEM VIII. • To find the Area of a Trapezoid. **RULE *. Multiply half the sum of the two parallel...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,... | |
| Matthew Iley - 1820
...Quadrilateral wherein two unequal Sides are Parallel to one another. RULE. Multiply half the sum of the **parallel sides by the perpendicular distance between them, and the product will be the area.** Let ABCD be a quadrilateral, wherein AC and BD are parallel but unequal; and let EF, the perpendicular... | |
| Anthony Nesbit, W. Little - 1822
...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... | |
| Abel Flint - 1825 - 241 sider
...and 8925 X 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.... | |
| Peter Nicholson - 1825 - 372 sider
...the arca of ABCD. MENSURATION. Prob. 6. To find the area of a trapezoid. Multiply the half sum of the **parallel sides by the perpendicular distance between them, and the product will be the area.** Ex. 3. What is the area of a board or plank in the form of a trapezoid, being If. 7¡- at one end,... | |
| John Nicholson - 1825 - 795 sider
...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... | |
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