Let length of rectangle = 5kand breadth of rectangle = 3kAccording to the quecation,5k - 3k = 8 ? 2k =8? k = 4? Lenght = 5k = 5 x 4 = 20 mBreadth = 3k = 3 x 4 = 12 m? Required area = Lenght x Breadth = 20 x 12 = 240 sq 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 %
Let the length, breadth and height of the room be l, b and h respectively As per question Cost of 2(l + b) x h = Rs. 48 ? Required cost = cost of 2 (2l + 2b) x 2h= cost of 4 [2(l + b) x h ]= 4 x Rs. 48= Rs. 192
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
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