If the area of a rectangle is $\sqrt{3} d^2$, where $2d$ is the length of its diagonal, then its perimeter is equal to

### Concept & Algebraic Identities in Geometry To solve problems involving the sum and product of two variables (length and breadth), use standard algebraic identities. Key formulas for a rectangle: $$Area = l \times b$$ $$Diagonal^2 = l^2 + b^2$$ We can bridge these using the identity: $$(l + b)^2 = l^2 + b^2 + 2lb$$ ### Step-by-Step Solution * **Given:** Area $lb = \sqrt{3} d^2$, and Diagonal $D = 2d$. * Calculate $D^2$: $D^2 = (2d)^2 = 4d^2$. So, $l^2 + b^2 = 4d^2$. * We need the perimeter, which is $2(l + b)$. First, find $(l + b)$. * Using the identity: $(l + b)^2 = (l^2 + b^2) + 2lb$. * Substitute the known values: $(l + b)^2 = 4d^2 + 2(\sqrt{3} d^2) = (4 + 2\sqrt{3}) d^2$. * Simplify the term $(4 + 2\sqrt{3})$ by expressing it as a perfect square: $3 + 1 + 2\sqrt{3} = (\sqrt{3})^2 + 1^2 + 2(\sqrt{3})(1) = (\sqrt{3} + 1)^2$. * Therefore, $(l + b)^2 = (\sqrt{3} + 1)^2 d^2$. * Taking the square root of both sides gives $l + b = (\sqrt{3} + 1) d$. * The perimeter is $2(l + b) = 2(\sqrt{3} + 1) d$. ### Exam Strategy & Shortcut Recognizing $4 + 2\sqrt{3}$ as $(\sqrt{3} + 1)^2$ is a frequent pattern in competitive exams. Memorizing basic surd expansions saves crucial time over trying to solve for $l$ and $b$ individually. ### Common Pitfall A common mistake is attempting to set up a quadratic equation to find the individual values of length and breadth. This approach is mathematically valid but incredibly time-consuming and prone to calculation errors. ### Final Answer Therefore, the correct answer is **$2(\sqrt{3} + 1) d$**.