 | Charles Hutton - Measurement - 1788 - 728 pages
...is 10-9956 yards? Anf. 86-19266. PROBLEM IX. *To fnd the Area of any Seftor of a Circle. RULE I. si Multiply the radius, or half the diameter, by half the arc of the feet or, and the product will be the area, as in the whole circle.* EXAMPLE I. What is the area of... | |
 | Mathematics - 1801 - 658 pages
...yards ? Ans. 86*19266. PROBLEM • PROBLEM XII. To find the ana of a sector of a circle. RULE •!.* _ 'Multiply the radius, or half the diameter, by half...the sector. NOTE. The arc may be found by Problem X. •c •"—" • RULE 2.f As 360 is to the degrees in the arc of the sector, so is the whole area... | |
 | Thomas Hodson - Education - 1806 - 576 pages
...CIRCLE. Fig.22. RULE i. Multiply the radius by half the arc •f the fe&or, and the product is the area. RULE 2. As 360 is to the degrees in the arc of the fe£tor, fo is the. whole area of the circle to the area of the fedor. Nate. For the femieircle take... | |
 | Charles Hutton - Mathematics - 1807 - 464 pages
...circles are 10 and 20 ? Ans. 235'62. PROBLEM XI. • To fnd the Area of the Secto'r of a Ci ' RULE I. MULTIPLY the radius, or half the diameter, by half the' arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take ^ of the product. The reason of... | |
 | Samuel Webber - Mathematics - 1808 - 466 pages
...cumference being 10* 9956 yards ? PR0BLEM XII. Tofind the area of a sector of a circle. RULE I.* .. Multiply the radius, or half the diameter, by half the arc of the sector for the area. Or, take i of the product of the diameter and arc of the sector. N0TE. The arc may be found by Problem X. RULE... | |
 | Charles Hutton - Mathematics - 1811 - 494 pages
...circles are 10 and 20 ? Ans. 235-62. PROBLEM XI. To find the Area of the Sector of a Circle. RULE I. MULTIPLY the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take -f .of the product. The reason... | |
 | Charles Hutton - Mathematics - 1811 - 442 pages
...rule to problem 9, for the whole circle. RULE II. Compute the area of the whole circle : then say, as 360 is to the degrees in the arc of the sector, so is the area of the whole circle, to the area of the sector. This is evident, because the sector is... | |
 | Charles Hutton - Arithmetic - 1818 - 646 pages
...236- 62. PBOBtEM XL To find the Area uf the Sector of a Circle. RULE l. MULT1PLY the radius, or holt the diameter, by half the -arc of the sector, for the area. Or, multiply the whole diameter by the whole ITC. of the sector, and take i of the product. The ^esron... | |
 | Charles Hutton - Mathematics - 1822 - 616 pages
...circles are 10 and 20 ? Ans. 235-62. PROBLEM XI. To find the Area of the Sector of a Circle. RULE I. MULTIPLY the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take J of the product. The reason of... | |
 | Anthony Nesbit - Surveying - 1824 - 476 pages
...chord of half the arc 45 feet, and the radius 37 feet 6 inches? Ans. 2.617ft. 10 in. 6 pa, RULE II. As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector. Kate I. The area of a semicircle, a quadrant,... | |
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MULTIPLY the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take -f .of the product. 
