Area Problems
The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
Let lateral side = (5y) cm and base = (4y) cm ? perimeter = 5y + 5y + 4y = 14 ?y = 1So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm? Required Area = ? (7 x 2 x 2 x 3) cm2=2?21 cm2
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
Increase in circumference of circle = 5%? Increase in radius is also 5%.Now, increase in area of circle = 2a + (a2/100) %Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%
EXAMIANS