Area Problems
A room of 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor.
area of the room = 544 x 374 sq.cm size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm area of 1 tile = 34x34 sq cm no. of tiles required = (544 x 374) / (34 x 34) = 176
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%
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
Diagonal of square = ?2a [a = side]4?2 = ?2 a a = 4 cmNow, area of square = a2 = (42) = 16Side of a square whose area is 2 x 16.a12 = 32 ? a1 = ?32 ?a14?2Now, diagonal of new square = ?2a = ?2x 4 ?2 = 8 cm
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
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%
EXAMIANS