Area Problems
A rectangle has 30 cm as its length and 720 sq cm as its area. Its area is increased to 1 1/ 4 times its original area by increasing only its length. Its new perimeter is
Original breadth of rectangle = 720/30 = 24 cmNow , area of rectangle = (5/4) x 720 = 900 cm2? New length of rectangle = 900/24 = 37.5 cm? New perimeter of rectangle = 2(l+ b)= 2(37.5 + 24 ) = 2 x 61.5 = 123 cm
Area of 4 walls = 2(l+b)h=2(10+7) x 5 = 170 sq mArea of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq marea to be planted = 170 -12 = 158 sq m Cost of painting = Rs. 158 x 3 = Rs. 474
Area of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900
Let length = L and breadth = BLet , New breadth = ZThen, New length = ( 160 / 100) L.= 8L / 5? 8L / 5 x Z = LBor Z = 5B/8Decrease in breadth = (B-5B/8)= 3B/8? Decrease in percent = (3B/8 x1/B ) x 100 %= 371/2%
EXAMIANS