The area of a square is twice that of a rectangle. The perimeter of the rectangle is 10 cm. If its length and breadth each is increased by 1 cm, the area of the rectangle become equal to the area of the square. The length of side of the square is :

Answer: Option A

Let the length and breadth of the rectangle be $$l$$ cm and n cm respectively
Then,
$$\begin{array}{rl}& 2\left(l+b\right)=10\\ & \Rightarrow l+b=5\\ & \Rightarrow b=\left(5-l\right)cm\end{array}$$
Area of the rectangle :
$$\begin{array}{rl}& =l\left(5-l\right)c{m}^2\\ & =\left(5l-l^2\right)c{m}^2\end{array}$$
Area of the square :
$$\begin{array}{rl}& =2\left(5l-l^2\right)c{m}^2\\ & =\left(10l-2l^2\right)c{m}^2\end{array}$$
$$\begin{array}{rl}& \therefore \left(l+1\right)\left(6-l\right)=\left(10l-2l^2\right)\\ & \Rightarrow l^2-5l+6=0\\ & \Rightarrow \left(l-3\right)\left(l-2\right)=0\\ & \Rightarrow l=3\end{array}$$
Area of the square :
$$\begin{array}{rl}& =\left(10\times 3-2\times 9\right)c{m}^2\\ & =12c{m}^2\end{array}$$
∴ Side of the square $$=\sqrt{12}cm=2\sqrt{3}$$