• # question_answer The perimeters of both, a square and a rectangle are each equal to 48 m and the difference between their areas is $4\,\,{{m}^{2}}$. The breadth of the rectangle is A)  10 m                B)  12 m    C)  14 m                D)  None of the above

Let the length of rectangle $=x\,m$ and              its breadth = y m Also, let the side of the square be z m. Then,       $2(x+y)=4z=48$ $\Rightarrow$               $x+y=24$ and $z=12$ Also,                 ${{z}^{2}}-xy=4$ $\Rightarrow$               $xy={{z}^{2}}-4=144-4=140$ So,                   ${{(x-y)}^{2}}={{(x+y)}^{2}}-4\,xy$ $=576-560=16$ $\therefore$      $x-y=4$ and $x+y=24$ On solving the above equations, we get $2y=20$ $\therefore$      $y=10\,m$