Area Problems
The area of a trapezium is 384 sq.cm 2. If its parallel sides are in ratio 3 : 5 and the perpendicular distance between them be 12 cm, The smaller of parallel sides is?
Let the sides of trapezium be 5k and 3k, respectively According to the question, (1/2) x [(5k + 3k) x 12] = 384? 8k = (384 x 2)/12 = 64 ? k = 64/8 = 8 cmLength of smaller of the parallel sides = 8 x 3 = 24 cm
Area of the room =(544 x 374) cm2size of largest square tile = H.C.F. of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) cm2? Least number of tiles required = (544 x 374) / (34 x 34) = 176
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 metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%