16 MENSURATION (5) When the four sides are given and the opposite angles are supplementary. In fig. 8 let AB = BC = Z>, CD = c, DA = d, AC = A, angle ABC = o, angle A. CDA = j8, and let a + = 180, then area of ABCD = ABC + ADC But sin j9 = sin (180 - a) = sin a, therefore area of ABCD = ^(ab + cd} sin a . This gives us the area of the Fig. 8. quadrilateral in terms of the four sides and one angle. Again we have 2 + b 2 - 2ab cos a = 7t 2 = c 2 + cP - 2cd cos = c 2 + d 2 + 2cdcos a , therefore 2(oi + cd) , and hence (a + b + c - d)(a + b-c + d) l + cosa = ,../. ^ - , and 2(ab + cd) 1 -cosa = From this we obtain 2(ab + cd) sin 2 a = (l+coso)(l -cos a) (6 + c + d - a)(c + d + a-b)(d + a + b-c)(a + b + c- Now let then (ab + cd) sin a = V(s - a)(s - b)(s - c)(s - d) ; therefore area of ABCD = V(s - a)(s - b)(s - c)(s - d) , or log area = j{log(s - rt) + log (s - J) + log (s-c) + log(s - d)} . If d = 0, the quadrilateral becomes a triangle, and its area is V (5 - a)(s - b)(s - c)s as before. In extracting the square root of sin a we take the positive sign, since the angle o is less than two right angles. G. Regular Polygons. 15. Since a regular polygon is both equilateral and equiangular, a circle can be inscribed within it and also described about it, and thus the n straight lines drawn from the common centre of the two circles to the n vertices of the polygon divide it into n triangles equal in every respect. Therefore the area of the polygon is equal to n times the area of any one of these triangles. 16. Radius of Inscribed and Circumscribed Circles. Let AB (fig. Q) = a be a side of a regular polygon of n sides ; let C be the centre of the inscribed nnd circumscribed circles, CD = ?- the radius of the former, and CE-R the radius of the B E Fig. 9. latter. The angle ACB is evidently equal to the ?ith part of four right angles, that is 360 C ACB = : and AD = - CD tan ACD = rtan - , 2 n 18_0 n 180 r = a x J cot Now and therefore and 17. Perimeter of Polygon. The perimeter of the polygon of 1 80 1 80 n sides is na, i.e., 2rtan , or 2/iRsin n n From this it follows that the perimeters of the inscribed and circum scribed regular polygons of n sides of a circle of radius r are . 180 180 2?irsm and 2/i?-tan respectively. 18. Area of Polygon. (a) In terms ofr. The area of polygon 180 ^60 n
(0) In terms of R. The triangle ACB = iAC.CBsinACB = i,R 2 and therefore area of polygon = ^?iR 2 sin 76 (7) In terms o/a. The triangle ACB i *T> OT> a a ,180 a 2 t 180 = *A13 . 01) = x / = - x Aacot = cot 22 n 4 n and therefore area of polygon = 2 x cot , From a and it follows that the areas of the inscribed and circum scribed regular polygons of n sides of a circle of radius r are Ar 2 sin --- and ?w 2 tan - respectively. n n 19. In the formula ( 18, 7) for the area of a polygon, the factor cot - has a definite value for every value of n, and hence, 4 n if we find its value once for all for a large number of values of n, and tabulate the results, we can find the area of a regular polygon of n sides by multiplying the square of its side by the appropriate tabular value. Again, if a = l we have r = i cot L and R = i cosec *- n n and thus we obtain in a similar manner the radius of the inscribed and circumscribed circles by multiplying the side by the appro priate tabular value of ^cot - and cosec --- respectively. 20. The following table contains the values of cot 180 and 1 80 1 80 their logarithms, and the values of icot and i cosec for all n n values of n from 3 to 12. n ^-cot 1 -^ 4 n Logarithms. .180 icot
icosec 1 -^ n 3 0-4330127 T-63C500fi 0-28867 57735 4 1-0000000 o-ooooooo 0-50000 70710 5 1-7204774 0-2350490 C-68819 85065 6 2-5980762 0-4146519 0-8(i602 1-0000 7 3-6339124 0-5603745 1-0383 1-1523 8 4-8284271 0-6380568 1-2071 1-3065 9 6-1818242 0-7911166 1-3737 1-4619 10 7-6942088 0-8861640 1-5388 1-6180 11 9-3656407 0-9715375 1-7028 17747 12 11-1961524 1-0490688 1-8600 19318 Let A denote the area of a polygon of n sides and A the cor responding tabular value of cot , then 4 n A = a- A , H. Length of the Radius of the Inscribed, Escribed, and Circumscribed Circles of a Triangle. 21. (a) Radhis of Inscribed Circle. Let O (fig. 10) be the centre of the circle inscribed in the triangle ABC and touching the sides in D, E, and F. Join OA, OB, and OC. The angles at D, E, F are right angles (Eucl. iii. 18). Let BC = rc, CA = b, AB = c, and OD = OE = OF = r. Now A B C = B C +COA+AOB = ar + ljb whence area of ABC Fig. 10. = (a + b + c)r = rs ; Vs(s -a)(s- b)(s - c) s s () Radius of Escribed Circles. Let OD = OE = OF = ?- a , then ABC = ACO + ABO - BOG = l>br n + ^cr a ^ar a = ^(b + c- a)r a r a (s - a) , area of ABC /s(s-a)(s-b)(s-c) /s(s-b)(s-c) and r a - = A/ " a" Similarly s(s-a)(s-c) - ;
s(s-a)(s- b)
