 | Peter Nicholson - 1809 - 426 σελίδες
...51208 61(5*26752 the «r*a of the circle. .PROBLEM XVII. To find the area of a sector of a circle. Multiply the radius, or half the diameter, by half...arc of the sector, and the product will be the area. EXAMPLE I. JHtat M the area of a sector ABC, the arc BC being 3f. 6i. and the radnu AB or AC 6f. 2i.... | |
 | Peter Nicholson - 1825 - 1046 σελίδες
...69696 54208 616 26732 the area of the tírele. P ral. 1/. To find the area of a sector of a circle. Multiply the radius, or half the diameter, by half...arc of the sector, and the product will be the area. Ex. \ . What is the area of a sector ЛВС, the arc BC being 3f. 6i. and the radius AB or ЛС 6f.... | |
 | John Nicholson - 1825 - 838 σελίδες
...19 2 88 : 7 : : 986'96044 7 8 I 690872308 11 I 863-59038 78-50821 Prol. 1 2. To find the Area of the Sector of a Circle. Rule\. Multiply the radius, or half the diameter, by half the arc of the sector, for the area. Or take ^ of the product of the diameter and arc of the sector. Note.... | |
 | Samuel YOUNG (of Manchester.) - 1833 - 272 σελίδες
...^' (" a fourth proportional I which being added to I the segments' height will ^ give the diameter. PROBLEM XI. To find the area of a Sector of a Circle. RULE 1. Multiply the length of the arc by the radius, and half the Product will be the area. RULE 2. cause.... | |
 | 1837 - 800 σελίδες
...instance, versed sine =18 and chord 48, then I83 =607- and *KX1«X2 48 X * C0'7 = 0367 the area nearly. To find the Area of a Sector of a Circle. Rule.— Multiply the length of the arc by the radius of the circle, and half the product will be the area. Example, —... | |
 | Frederick Walter Simms - 1845 - 184 σελίδες
...subtract the chord of the whole arc ; one third of the remainder will be the length of the arc, nearly. To find the area of a sector of a circle. Rule. — Multiply the length of the arc by half the length of the radius, the product will be the area. To find the area... | |
 | William Templeton (engineer.) - 1846 - 236 σελίδες
...diameter. 355 Or, V 122.71875 = 11.077; and 11.077 X 1.12837 = 12.49895, or 12.5 diameter. PROBLEM X. To find the area of a sector of a circle. RULE. — Multiply the length of the arc by the radius of the circle, and half the product will be the area. EXAMPLE. —... | |
 | Peter Nicholson, Joseph Gwilt - 1848 - 750 σελίδες
...616-26752 the area of the circle. PROBLEM XVII. PLATE 51. To ßnd the area of a sector of a circle. Multiply the radius, or half the diameter, by half...arc of the sector, and the product will be the area. EXAMPLE. What is the area of a sector ABC, the arc вс being 3f. 6i, and the radius AB or AC Of. 2i... | |
 | Charles Haslett - 1855 - 482 σελίδες
...12 "5, diameter. 355 Or, V122-71875 = 11-077 ; and 11-077 x 1-12837 = 12-49895, or 12'5, diameter. PROBLEM XI. To find the area of a sector of a circle. RULE. Multiply the length of the arc by the radius of the circle, and half the product will be the area. EXAMPLE. Required... | |
 | Charles Haslett - 1855 - 544 σελίδες
...= 12-5, diameter. 355 Or, ^122-71875 = 11-077 ; and 11*077 x 1-12837 = 12*49895, or 12-5, diameter. PROBLEM XI. To find the area of a sector of a circle. RULE. Multiply the length of the arc by the radius of the circle, and half the product will be the area. EXAMPLE. Required... | |
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Problem 14. — To find the area of a sector of a circle. Rule. — Multiply the arc of the sector by half the radius. Example. — How much tin is required to cover a 60° sector of a 10 foot circular deck? 
