Area Problems
A rectangle has 30 cm as its length and 720 sq cm as its area. Its area is increased to 1 1/ 4 times its original area by increasing only its length. Its new perimeter is
Original breadth of rectangle = 720/30 = 24 cmNow , area of rectangle = (5/4) x 720 = 900 cm2? New length of rectangle = 900/24 = 37.5 cm? New perimeter of rectangle = 2(l+ b)= 2(37.5 + 24 ) = 2 x 61.5 = 123 cm
Area of the room=(544 * 374) size of largest square tile= H.C.F of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) c m 2 Number of tiles required== [(544 x 374) / (34 x 34)] = 176
Area of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
EXAMIANS