Original area = (22/7) x 9 x 9 cm2New area = (22/7) x 7 x 7 cm2? Decrease = 22/7 x [(9)2 -(7)2] cm2=(22/7) x 16 x 2 cm2Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %= 39.5 %
Let length of rectangle = 5kand breadth of rectangle = 3kAccording to the quecation,5k - 3k = 8 ? 2k =8? k = 4? Lenght = 5k = 5 x 4 = 20 mBreadth = 3k = 3 x 4 = 12 m? Required area = Lenght x Breadth = 20 x 12 = 240 sq m
Let length of the rectangular field = 7k m and breadth of the rectangular field = 2k mAccording to the question,Area of a rectangular field = Length x Breadth? 3584 = 7k x 2k ? 14 x k2 = 3584 ? k2 = 3584/14 = 256? k2 = 256 = 16 m? Length of rectangular field = 7k = 7 x 16 = 112 mAnd breadth of rectangular field = 2 x 16 = 32 m? Perimeter of rectangle = 2(Length x Breadth)= 2(112 + 32) = 2 x 144 = 288 m
EXAMIANS