Area Problems
The breadth of a rectangle is 25 m. The total cost of putting a grass bed on this field was ? 12375, at the rate of ? 15 per sq m. What is the length of the rectangular field?
Area to the rectangular field = 12375/15 = 825 sq mAccording to the question, (L x B) = 825 [L = length and B = breadth]? L x 25 = 825 ? L = 825/25 = 33 m
Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2) = 2(15 + 2) = 34 cm
Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
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
EXAMIANS