| Mathematics - 1801
...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 - Geometry - 1804 - 216 pages
...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 - Geometry - 1808 - 192 pages
...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 pages
...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 - Surveying - 1813 - 168 pages
...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 pages
...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 - 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... | |
| Abel Flint - Geometry - 1825 - 252 pages
...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 - Mathematics - 1825 - 372 pages
...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 - 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... | |
| |