Area Problems
The length and breadth of a square are increased by 40% and 30% respectively. The area of a resulting rectangle exceeds the area of the square by?
Let the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
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
? 22/7 x r2 = 13.86 x 10000? r2 = (13.86 x 10000 x 7) / 22? r = 210 m ? Circumference = [2 x (22/7) x 210 ] m = 1320 m Cost of fencing = Rs. (1320 x 20)/100= Rs . 264
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