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%
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
Given that, l = 2b [Here l = length and b = breadth]Decrease in length = Half of the 10 cm = 10/2 = 5 cmIncrease in breadth = Half of the 10 cm = 10/2 = 5 cm Increase in the area = (70 + 5) = 75 sq cm According to the question, (l - 5) (b + 5) = lb + 75 ? (2b - 5) (b + 5) = 2b2 + 75 [since l = 2b]? 5b - 25 = 75 ? 5b = 100? b = 100/ 5 = 20? l = 2b = 2 x 20 = 40 cm
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
EXAMIANS