Area Problems
A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is
Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
Let the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
According to the question, Area of semi- circle = 77 m(1/2) x ? x r2 = 77? r2 = (77 x 2 x 7)/22? r = 7m ?Circumference of semi- circle =?r + 2r= ( ? + 2)r = [(22/7) + 2] x 7 = 36 m
Perimeter = Distance covered in 8 min. = ❨ 12000 x 8 ❩m = 1600 m. 60 Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. ∴ Length = 480 m and Breadth = 320 m. ∴ Area = (480 x 320) m2 = 153600 m2
Let original radius be r.Then, according to the questions,? (r + 1)2 - ?r2 = 22? ? x [(r + 1)2 - r2] = 22? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22? 2r + 1 = 7 ? 2r = 6 ? r = 6/2 = 3 cm
EXAMIANS