Page:Encyclopædia Britannica, Ninth Edition, v. 16.djvu/28

Rh 18 MENSURATION As an example of the use of this table we proceed to find the value of when x = 68 45 17&quot; 8. When x =68 = 1-1868239, a;= 5 0= -0014544, x = 10&quot; 0= -0000485, x = 7&quot; 0= -0000339, and when x = 0&quot; -80= &quot;0000039 , therefore when z = 68 45 17&quot; 8 = 1-2000001. 30. Combining the results of 27 and 28 we obtain s 180 s. (a) =, and=-. , T IT r s _ 180 _s_. TT x (J8) (7) _ 31. Length of Arcs of Circles. The following are the more im portant cases: (a) In terms of the chord of the arc and the radius of the circle. Let AB (fig. 14) = 2c, AC = r, and AEB=s, then AD = JAB = c=rsin^C, whence C is known, and therefore the arc s is found ( 30, 7). (/3) In terms of the height of the arc and the radius of the circle. Let DE = A = height of arc, then CD-CE-DE-r-A, in CD r-h and cosJC = = , AC r whence C is found, and therefore s. B E Fig. 14. 32. Huygens s Approximation to the Length of a Circular Arc. Let AB (fig. 15) =p be the chord of the arc AEB, and AE = EB = g that of half the arc, then the arc AEB= J (Bq-p) approximately. For, let r denote the radius, s the arc AEB, and 20 the angle ACB, then #= nT Again, A.B=p and similarly q = 2r sin Now = 2rsin0 = 2rsin 4r sin = 3 5 - .-=- + nr - &e. |D therefore ,-S, .,-t- 2 . 3 . 4 . r- 2. 3 . 4 . 5 . 16 . I l/ 1 / s ) Similarly 8g = l6r ~4- +nr( T- ) - *e. ( 14?- |3)-/ |5r/ ) 2 . 3 . 4 . ?* 2 . 3 . 4 . 5 . 64 . ,-&c. Hence, neglecting powers of be- r yond the fourth, we obtain approximately. In practice it is sometimes more convenient to use the equivalent form 33. Area of Sector of a Circle. Let the sector be OAB (fig. 16). Divide the arc AB into n equal parts, and draw the chords of these. Let P denote the perimeter of the broken line AB, A the area of the polygon AOB, and p the alti- Fig. 16. tude of any one of the n equal triangles of which this polygon is made up. Now in the limit, when n is indefinitely increased, P becomes the arc AB = s, a result which we symbolize thus L P = arcAB = s. ~^H=X Similarly L _ _p = radius = r. Again, the area of the sector is equal to the area of the polygon when the broken line AB becomes the arc AB, that is, sector = P. 34. Let S denote the area of a sector of a circle, then, by means. of the above result and 27, we have (a) S = Jsr, 03) S = 4r0. r = r-0. 35. We proceed to find the area of a sector of a circle in the fol lowing additional cases : (a) When the chord of the sector and the radius of the circle are given. In fig. 14 let AB = 2c, and let AC = r, then we have . ACB AD c j i r _ _ _ _ 2 AC r whence ACB and therefore is known, and S can be found by 34. ACB has two values, the smaller one giving the area of the minor, and the larger that of the major conjugate sector. (0) When the. chord and height of the chord are given. Let DE (fig. 14) =7; and AB = 2c, then AC 2 = r 2 = AD 2 + DC 2 = c- + (r - h}~, whence
 * 1) = 40 0= -0116355,

and therefore by previous case the area can be found. (7) Wh-f.n the, chord and angle subtended at the centre are given. Let AB (fig. 14) =2c and ACB = 0, then c. ACB c = sin, or r = ^ , therefore area of sector = r-6 = X0. .ship/ 36. Area of a Circle. The circle being a sector whose arc is the- whole circumference we obtain at once area of circle = ^r x s *= ^r x 2?rr = irr-. An independent proof of this proposition might be given by means of the inscribed and circumscribed polygons, and from the area of a circle the area of a sector can be deduced. The infini tesimal calculus affords a simple and elegant proof (see 44). 37. If A denote the area, r the radius, d = 2r the diameter, and C the circumference of a circle, we have (a) C 2 ^ nd- 4TT 2 7 2 ^TT~&quot; (7) (8) t. Whence we see that the area of a circle is obtained by multiplying, (a) the square of its radius by 7r = 3 14159, (j8) the radius by half the circumference, (7) the square of the circumference by = 07957 , (5) the square of the diameter by TT= 78539. 38. Again, from the above formulae we deduce T- = -4- A = 5641 896 x A, Vir d-JL = A-l 1283792xA, (a) (0) (7) thus obtaining radius, diameter, and circumference from area. 39. Area of a Circular Ring. Let r and r l denote the radii of the outer and inner circles respect ively (fig. 17), then the area of the space between them = irr 2 - irr- = ir(r + rj(r - rj. The circles need not be concentric, and the reader should note that the area of the ring is equal to the area of an ellipse whose major and minor axes are r + i and rr^ (see 51). 40. Area of the Sector of an Annulus. Let angle ACB = in fig. 17, then the area of ABED = sector ACB - sector DCE Fig. 17. E