 | Peter Nicholson - 1809 - 426 páginas
...189 343) 1200 1029 17100 PROBLEM V. To find the area of a trapezium. Multiply the diagonal by half the sum of the two perpendiculars, falling upon it from the opposite angles, and the product will be the area. EXAMPLE I. What is the area of a trapezium ABCD, the diagonal AC being... | |
 | Anthony Nesbit - 1824 - 476 páginas
...areaffa trapezium. RULE. Multiply the sum of the two perpendiculars by the diagonal upon which they fall, from the opposite angles, and half the product will be the area. Or, divide the trapezium into two triangles, in the most convenient manner ; and the sum of their areas... | |
 | Peter Nicholson - 1825 - 1046 páginas
...18 the 18 144 18 324 17100 Prob. 5. To find the area of a trapezium. Multiply the diagonal by half the sum of the two perpendiculars-, falling upon it from the opposite angles, and the product will be the area. Ex. What is the area of a trapezium ABCD, the diagonal AC being 36 feet,... | |
 | Thomas Hornby (land surveyor.) - 1827 - 318 páginas
...sides~1440 400 2)576000 Double area. 2.88000 4 3.52000 40 20.80000 • Ans. ZA. 3n. 20p. PROBLEM 4. , .. To find the Area of a Trapezium. RULE. Multiply the diagonal by the sum of the two perpendiculars, let fall upon it, from its opposite angles ; and half the product will be the area. EXAMPLE. Required... | |
 | John Bonnycastle - 1829 - 256 páginas
...the other side ; required the breadth of the street. Ans. 76. 1233 feet . OF SUPERFICIES. PROBLEM V. To find the area of a trapezium. RULE.* Multiply the...and half the product will be the area. EXAMPLES. 1. Required the area of the trapezium BAED, whose diagonal BE is 84, the perpendicular AC 21, and DF 28.... | |
 | Thomas Curtis - 1829 - 810 páginas
...separately, and their sum will be the area of the trapezium ; or multiply the diagonal by half the sum of the perpendiculars falling upon it from the opposite angles, and half the product will be the area. Example. — What is the area of a trapezium whose diagonal is 80-5, and perpendiculars 24-5, and 30-1... | |
 | John Bonnycastle - 1833 - 310 páginas
...window, 36 feet high, on the other side ; required the breadth of the street. Ans. 76.1233/eei PROBLEM V. To find the area of a trapezium. RULE.* Multiply the diagonal by the sum of the two perpendicu lars falling upon it from the opposite angles, and half the product will be the area. EXAMPLES.... | |
 | Charles Haynes Haswell - 1844 - 298 páginas
...TRAPEZIUMS AND TRAPEZOIDS. To find the Area of a Trapezium — fig. 8. RULE. — Multiply the diagonal ac by the sum of the two perpendiculars falling upon...opposite angles, and half the product will be the area. To find the Area of a, Trapezoid — fig. 9. RULE. — Multiply the sum of the parallel sides a 4,... | |
 | John Bonnycastle - 1848 - 312 páginas
...window, 36 feet high, on the other side ; required the breadth of the^treet. Ans. 76.1233/eet PROBLEM V. To find the area of a trapezium. RULE.* Multiply the diagonal by the sum of the two perpendicu lars falling upon it from the opposite angles, and half the product will be the area. ,... | |
 | John Bonnycastle - 1848 - 320 páginas
...To find the area of a trape2ium, RULE.* Multiply the diagonal by the sum of the two perpendieu lars falling upon it from the opposite angles, and half the product will be the area. EXAMPLES. 1. Required the area of the trapezium BAED, whose diagonal BE is 84, the perpendicular AC 21, and DF 28.... | |
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To find the area of a trapezium. RULE. — Multiply the diagonal by the sum of the two perpendiculars falling ?;„ upon it from the opposite angles, and divide the product by 2. 
