Area Problems
The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle is increased by 75 cm 2 . Therefore , the length of the rectangle is?
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
According to the question,l/[2(l + b)] = 5/18? 10l + 10b = 18l = 10b? l/b = 10/8 = 5/4? l : b = 5 : 4Hence, ratio of length and breadth of a rectangle is 5 : 4
Let the diagonal of one square be (2d) cmThen, diagonal of another square = d cm? Area of first square = [ 1/2 x (2d)2] cm2Area of second square = (1/2 x d2) cm2? Ratio of area = (2d)2/ d2= 4/1 = 4: 1
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