 | Mathematics - 1801 - 658 pages
...the 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...between them, and half the product will be the area. • EXAMPLES. * DEMONSTRATION. or (because B»=DE) =-, .-. A ABD+ A BCD, or , , , ABxDE , DCxDE the... | |
 | James Thompson - Arithmetic - 1808 - 176 pages
...trafiezoid, or quadrangle, <u'o cf •whose opposite sides are parallel. RULE — Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product •will be the area. EXAMPLES. 13. Required the area of a trapezoid whose parallel sidef »re 25 feel 6 inches, and 18 feet... | |
 | Anthony Nesbit, W. Little - Measurement - 1822 - 916 pages
...the grain? 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...between them, and half the product will be the area in square inches. Divide this area by 2 82, 231, and 2150.42, and the respective quotients will be... | |
 | Anthony Nesbit - Surveying - 1824 - 476 pages
...Ans. 1131^.2 in. 9 pa. PROBLEM VIII. To find the area of a Irapezoid. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. Or, half the sum of the sides multiplied by their distance will give the area. EXAMPLE. 1. What is... | |
 | John Nicholson - Machinery - 1825 - 838 pages
...when the squre of AB has been subtracted. 63 I 189 3 1 189 Prot. 4. To find the Area ofaTrapezoid. Multiply the sum of the two parallel sides by the...perpendicular distance AP or CN is 9 : required the area. 7 12 19 9 171 85£ the area. Prob. 5. To find the Area of a Trapezium. Case for any trapezium. — Divide... | |
 | John Nicholson (civil engineer.) - Great Britain - 1825 - 1008 pages
...Prob. 4. To find the Area of aTrapezoid. Multiply the sum of the two parallel sides by the perpendini distance between them, and half the product will be...In a trapezoid, the parallel sides are AB 7, and CD I: and the perpendicular distance AP or CN is 9 : required the area, 7 12 19 9 171 85$ the area. Pfot.... | |
 | Thomas Hornby (land surveyor.) - Surveying - 1827 - 318 pages
...16.00000 Ans. 0A. 2n. 16p. PROBLEM 3. To find the Area of a Trapezoid. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. EXAMPLE. Required the area of the trapezoid AB CD, whose parallel sides AB and DC are 800 and 640 links... | |
 | John Bonnycastle - Geometry - 1829 - 256 pages
...trapezoid, or a quadrangle, two of whose opposite sides are parallel. RULE.* Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the EXAMPLES. 1. Required the area of the trapezoid ABCD, whose sides AB and DC are 321.51 and 214.24,... | |
 | John Nicholson (Civil engineer) - Building - 1830 - 240 pages
...when the 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 AP or CN is 9 : required the area. 7 12 19 9 171 855 the area. Prob. 5. To find the Area of a Trapezium. Case for any trapezium.— Divide... | |
 | Edinburgh encyclopaedia - 1830 - 856 pages
...quadrilateral, of which two opposite sides are parallel but not equal. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product is the area. In the trapezoid ABCD, draw the diagonal AC, and from its extremities draw AE, CF at right... | |
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Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. 
