Area Problems
The dimensions of a room are 10m x 7m x 5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m x 3m. One window is of size 2m x 1.5m and the other 2 windows are of size 1m x 1.5m. The cost of painting the walls at Rs. 3 per sq m is
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
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%
speed = 12 km/h = 12 × 5 18 = 10 3 m / s distance covered = 20 × 2 × 22 7 × 50 = 44000 7 m time taken = distance /speed = 44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n
Area of square plate = (Side)2 = (2d)2= 4d2Area of circular plate = ? (d/2)2= ?d2/4? Number of square plates = [(4d2)/4] / [(?d2)/4]= (4 x 4)/? ? 5Since, nearest integer value is 5.
Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %
EXAMIANS