Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
Let a = 13, b = 14 and c = 15. Then, s = 1 2 a + b + c =21 (s- a) = 8, (s - b) = 7 and (s - c) = 6. Area = s s - a s - b s - c = 21 × 8 × 7 × 6 = 84 sq.cm
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
EXAMIANS