 | 1801 - 658 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 - 226 sider
...Angle 28° 5'. 0.47076 105X85=8925, and 8925X0.47076=4201 the double Area of the Triangle. PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half...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 - 190 sider
...28* 5'. 0.47076 105x85=8925, and 8925x0.47076=4201 the double Area of the Triangle. • PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half...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 - 426 sider
...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, 2f. 3i.... | |
 | Abel Flint - 1813 - 214 sider
...and 8925X0.47076=4201 the double Area of the Triangle. PROBLEM X. To find the Area of a Trapezovl. 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.... | |
 | Matthew Iley - 1820 - 512 sider
...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 - 916 sider
...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 - 252 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 - 1046 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 - 822 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... | |
| |
To find the area of a trapezoid. RULE. Multiply half the sum of the two parallel sides by the perpendicular distance between them : the product will be the area. 
