The areas of a square and a rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 5 cm and the breadth is less by 3 cm. Find the perimeter of the rectangle ?
Correct Answer: 34 cm
Let the length of each side of the square be x cm
Then, length of rectangle = (x + 5) cm and its breadth = (x - 3) cm
$$\eqalign{ & \therefore \left( {x + 5} \right)\left( {x - 3} \right) = {x^2} \cr & \Rightarrow {x^2} + 2x - 15 = {x^2} \cr & \Rightarrow x = \frac{{15}}{2} \cr} $$
∴ Length :
$$\eqalign{ & = \left( {\frac{{15}}{2} + 5} \right)cm \cr & = \frac{{25}}{2}cm \cr} $$
Breadth :
$$\eqalign{ & = \left( {\frac{{15}}{2} - 3} \right)cm \cr & = \frac{9}{2}cm \cr} $$
Hence, perimeter :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = 2\left( {\frac{{25}}{2} + \frac{9}{2}} \right)cm \cr & = 34\,cm \cr} $$
Then, length of rectangle = (x + 5) cm and its breadth = (x - 3) cm
$$\eqalign{ & \therefore \left( {x + 5} \right)\left( {x - 3} \right) = {x^2} \cr & \Rightarrow {x^2} + 2x - 15 = {x^2} \cr & \Rightarrow x = \frac{{15}}{2} \cr} $$
∴ Length :
$$\eqalign{ & = \left( {\frac{{15}}{2} + 5} \right)cm \cr & = \frac{{25}}{2}cm \cr} $$
Breadth :
$$\eqalign{ & = \left( {\frac{{15}}{2} - 3} \right)cm \cr & = \frac{9}{2}cm \cr} $$
Hence, perimeter :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = 2\left( {\frac{{25}}{2} + \frac{9}{2}} \right)cm \cr & = 34\,cm \cr} $$