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 becomes equal to the area of the square. The length of side of the square is

### Concept & Systems of Equations This problem requires setting up a system of equations linking the areas and perimeters of two distinct geometric shapes. $$ \text{Area of Rectangle} = l \times b $$ $$ \text{Perimeter of Rectangle} = 2(l + b) $$ ### Step-by-Step Solution 1. Let the side of the square be $S$. Its area is $S^2$. 2. Let the length and breadth of the rectangle be $l$ and $b$. Its area is $lb$. 3. We are given: $S^2 = 2lb$ (Equation 1). 4. The perimeter of the rectangle is 10 cm: $2(l + b) = 10 \Rightarrow l + b = 5$ (Equation 2). 5. If length and breadth are increased by 1 cm, the new area equals the square's area: $(l + 1)(b + 1) = S^2$ 6. Expand the left side: $lb + l + b + 1 = S^2$. 7. Substitute $(l + b) = 5$ into the expanded equation: $lb + 5 + 1 = S^2 \Rightarrow lb + 6 = S^2$. 8. We already know from Equation 1 that $S^2 = 2lb$. Substitute this in: $lb + 6 = 2lb \Rightarrow lb = 6$. 9. Now plug $lb = 6$ back into Equation 1 to find $S$: $S^2 = 2(6) = 12$. 10. The length of the side of the square is $S = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ cm. ### Exam Strategy & Shortcut Instead of solving for individual $l$ and $b$, recognize the algebraic pattern in $(l+1)(b+1) = lb + (l+b) + 1$. Since $l+b$ is immediately known from half the perimeter ($5$), you can skip to $lb + 6 = \text{New Area} = \text{Square Area} = 2lb$, solving for $lb$ in one fluid mental step. ### Common Pitfall Trying to solve the quadratic equation to find exactly what $l$ and $b$ are ($l=3, b=2$). While correct, it wastes valuable time; you only need their product ($lb = 6$) to find the square's area. ### Final Answer Therefore, the correct answer is **$2\sqrt{3}$ cm**.