 | Abel Flint - 1804 - 226 páginas
...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... | |
 | Abel Flint - 1808 - 190 páginas
...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... | |
 | Thomas Keith - 1817 - 306 páginas
...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 sides by the perpendicular distance between them, and the product will be the area. Example 1. Let AB c D JE. be... | |
 | Abel Flint - 1825 - 252 páginas
...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... | |
 | Ira Wanzer - 1831 - 408 páginas
...triangle whose sides are 30, 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... | |
 | Zadock Thompson - 1832 - 186 páginas
...16, 11 and 10 rods ; what is the area ? Ans. 54.299 rods. 307. To find the area of a trapezoid. (65) RULE. — Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will be the area. 1. One of the two parallel sides... | |
 | Zadock Thompson - 1832 - 182 páginas
...16, 11 and 10 rods ; what is the area ? Ans. 54.299 rods. 307. To find the area of a trapezoid. (65) RULE.— Multiply half the sum of the two parallel sides by the perpendicular distance hetween them, and the product will he the area. , 1. One of the two parallel... | |
 | Ireland commissioners of nat. educ - 1834 - 370 páginas
...С 35, and B С 16 ; required its area ? Ans. 700-99. PROBLEM XIII. 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 give the area.* 1. Let A BCD be a trapezoid,... | |
 | Abel Flint - 1835 - 368 páginas
...of the angle 28° 5' 0.47076. 105x85=8925, and 8925x0.47076=4201 the double area. pf the triangle. 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... | |
 | George Willson - 1836 - 202 páginas
...-figure. The Fig. 7. I breadth of the rectangle is «qual to half the sum of tha two parallel sides. RULE. — Multiply half the sum of the two parallel sides, by the perpendicular distance between them. All the figures we have considered, have been referred either... | |
| |
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. 
