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 square field = 1 hectare = 10000 m2Side of the square = ? 10000 m = 100 mSide of another square field = 100 + 1 = 101 m? Required difference of area = [(101)2 - (100)2] m2=[(101 + 100 ) (101 - 100) ] m2= 201 m2
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
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