Area of park = 100 x 100 = 10000 m2Area of circular lawn = Area of park - area of park excluding circular lawn= 10000 - 8614 = 1386Now again area of circular lawn = (22/7) x r2 = 1386 m2? r2 = (1386 x 7) / 22= 63 x 7= 3 x 3 x 7 x 7? r = 21 m
Original circumference = 2?r New circumference = (150 /100) x 2 ?r = 3?r 2?R = 3?r? R = 3r/2 Original area = ?r2New area = ?R2= ?9r2 / 4 = 9?r2/4Increase in area = (9?r2/4 ) - (?r2)= (5/4) ?r2Req. increase per cent = [{(5/4) ?r2} / {?r2}] x 100 = 125 %
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3
Let the radius of circular field = r m.Speed of person in m/s = 30/60 = 1/2m/sAccording to the question,[(2?r) /(1/2)] - [(2r)/(1/2)] = 30? 4?r - 4r = 30? [4 x (22/7) - 4]r =30? (125 - 4)r = 30 ? (8.5)r = 30? r = 30/8.5 = 3.5 m
Area to the rectangular field = 12375/15 = 825 sq mAccording to the question, (L x B) = 825 [L = length and B = breadth]? L x 25 = 825 ? L = 825/25 = 33 m