III. Or multiply the square of the circumference by ⚫079577. EXAMPLES. 1. What is the area of a circle whose diameter is 1 ? Half the diameter is 5, and half the circumference is 1.570796 &c. 5 x 1.570796785398 or 7854 nearly, the area. Now 7854 is a common multiplier for finding the area of any other circle whose diameter is given: thus, 2. Let it be required to find the area of a circle whose diameter is 20? Then circles being as the squares of their diameters (105, corol.) we have 12: 7854: 202: 400 X 7854 314 16 the area sought (rule II). 3. Required the area of a circle whose circumference is 1? As 3.1415926: 1 :: 1:31831 nearly, the diameter: Therefore X 31831 079577 the content. Which is a multiplier for finding the area when the circumference is given (Rule III.). 4. How many square yards in a circle whose radius is 15 feet? 5. Acre? Ans. 81.1798, nearly. What is the diameter of that circle whose area is an Ans. 78 yards, nearly. To find the area of the Sector of a Circle. 268. WHEN the diameter and length of the arc are given, Multiply half the diameter by half the arc, and the product will be the area: (this is evident from 106, corol.). EXAMPLES. 1. What is the area of the circular sector if the radius is 20, and length of the arc 36? 20 X 18369. Ans. 2. What is the area of a sector if the radius be 1, and the arc contains 40° ? When the radius is 1, the circumference is 6.2831853 (266): Therefore 6.2831853 01745 &c. length of the arc of 1°. And 01745 × 40 698 is the length of the arc of 40°. *698 And the area of the sector = 1 x 349 Ans. 269. Let CA, CG be the radii of two similar sectors CAD, CGO: Then CA: AD :: CG: GO, G D or 1 01745 X 40° :: CG: 01745 x 40° x CG the length of the arc GO when the angle C is 40°: Therefore if the number of degrees in a sector be multiplied by the radius and that product by the decimal 01745 the result will be the length of the arc of the sector. Since the area of the sector CAD is CA x AD, or 1 x 01745 X 40° 2 (if the angle C is 40°) it will be CA: CG2 :: 1 X CGO (105, corol.). 01745 x 40° : area of sector 2 or 1: CG :: 0087266 × 40°: CG2 x '0087266 x 40°: Consequently, if the square of the radius, the number of degrees in the sector, and the decimal 0087266 are multiplied together, the product will be the area. Examp. What is the area of a sector when the radius is 50, and its arc 94° 34'? 94° 34′1⁄2 94575 And 50 × 94-575 × 00872662063 2955 the area sought. 270. To find the area of a Segment of a Circle. Ex. 1. LET ADB be a segment whose chord AB = 36, and height or versed sine OD = 8; C being the centre of the circle. Now OB being 18, we get (224.) the angle OCB = 47° 55', therefore the angle of the sector ACB = 95° 51′ = 95°85. And the area of the sector ADBC 2. area of the triang. ACB= (OB×OC) Area of the segment ADB =491-88 (269.) =292.5 diff. 199.38 If the height or versed sine be 50, and the radius of the circle 40; what is the area of the segment? 271. A mixt-lined figure is one bounded by both right and curved lines, as AO. No general rule can be given for obtaining the exact contents of all figures of this description. The usual method of ap proximation is to divi le the curved or crooked lines into short parts, and then consider each of those parts as the side of a right-lined figure. Examp. 1. Suppose AD is divided into 3. equal parts, and let AE, BG, CI, DO, be perpendicular to AD; also suppose Then if we suppose EG, GI, 10, to be right lines, the figure will consist of 3 trapezoids having equal bases AB, BC, CD : But the same result is obtained by multiplying the arithmetical mean breadth by the base or length AD. Thus, take half the sum of the extreme breadths AE and DO for one breadth, to which add BG and CI, and divide the whole by 3 (the num ber of parts into which AD is divided) for the mean breadth. For the sum of the 3 fractions having the common denominator 2 is 2 Therefore the sum 7+ 8+ 10 x 7; or 25 × 21 251 x 21 1781, is equal to the 3 trapezoids: where 7} is half the extreme breadths AE, DO; and 7+9+10 or 84 is the mean breadth. It is evident that the area 49 is too great, because the curved side EG is convex towards the opposite side AB: but GI and IO are bent the contrary wav, and consequently 63 and 66 are the number both too little. Hence it appears, that the greater of equal parts into which the base (AD) is divided, the more accurate will be the result. Examp. 2. The length or base of an irregular figure being 37.6, and the breadths at 9 equi-distant places 0, 4·4, 6·5, 7·6, 5.4, 8, 52, 61, 6·5; what is the area? 272 Ans. 218 315. D The following method of reducing a crooked boundary to a straight line is sometimes practised in land-measuring. Suppose ABCD is a field protracted from a survey, the side DC being very iregular: Then to reduce this side to a straight line, lay a fine thread GR across it, and guess by the eye when the parts of the surface excluded on one side of the thread are equal to those taken in on the other; then draw the line GR with a pencil; and the surface of the field will be reduced to the quadrilateral ABRG. A fine silk thread, or horse hair, streth d after the manner of a bow-string, will be found very convenient for this purpose. Mensuration of Solids. 273. By the mensuration of solids we understand that of their superficies, as well as the capacities or solid contents. If a solid is bounded by planes they must be right-lined figures (125); and their areas added together will give the whole surface of the solid. |