The length of a rectangle is reduced by 20% and breadth

Solution

Let length of the rectangle = L, breadth of the rectangle = B Area= L*B Now, length reduced by 20%. So, now length – (4/5)L Breadth remains same So, new figure is square. So, (4/5)L = B So, New area = (4/5)L*B 1.  Difference in area = [{L/B – (4/5) L*B}/ L*B]*100 = 20% (So, wrong as given in the statement 1 to be 25%) 2.   Perimeter of square = 2(4/5) L + 2B = (8/5) L + 2B Perimeter of rectangle = 2(L + B) Difference = [{2L + 2B – (8/5)L – 2B}/ (2(L + B)]*100 = [(2/5)L/ 2(L + (4/5)L]*100 = 11.11% (So, right it is approx. 11%) 3.  Diagonal of rectangle = √(L² + B²) = √(L² + (4/5 L)²  = (L/5)*√41 Diagonal of square =side*√2 = (4/5)L*√2 Require   %age   =  [{(√41 - 4√2)*(L/5)}/(L/5)*√41]*100 = 11.65% (So, right it is approx. 12%) Hence option D.