Area Problems
A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by 1 1/ 5 time the original area by increase its length only then the perimeter of the rectangle so formed (in cm) is
Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
Length of carpet = Total Cost / Rate = 3600 / 30 = 120 mArea of carpet = (120 x 75) / 100 m2= 90 m2? Area of the room = 90 m2Breadth of the room = Area /Length = 90 / 15 m = 6m
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3
EXAMIANS