The sum of the perimeter of circle R and S together is 1012 cm. Let’s assume the radius of circle R and S is ‘r’ and ‘s’ respectively. 2x(22/7)x[r+s] = 1012 (44/7)x[r+s] = 1012 (1/7)x[r+s] = 23 r+s = 23x7 = 161 Eq.(i) If the area of circle S is 18634 cm2 . area of circle S = (22/7)x( s)2 = 18634 (1/7)x( s)2 = 847 ( s)2 = 5929 So s = 77 cm Put the value of ‘s’ in Eq.(i). r+77 = 161 r = 161-77 r = 84 cm Required percentage = (r/s)x100 = (84/77)x100 = (12/11)x100 = (1200/11) = 109% (approx.)
Statements: L > M = N > O, O > P ≥ T = R
Conclusion:
I. L > T
II. L ≥ R
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