Answer :
Answer:
MARK AS BRAINLIST
Step-by-step explanation:
To find the perimeter of the rectangle given that its area is \(\sqrt{3}d^2\) and \(2d\) is the length of its diagonal, we need to use the relationships between the dimensions of the rectangle.
Let the length and width of the rectangle be \(l\) and \(w\), respectively. The area \(A\) of the rectangle is given by:
\[ A = lw = \sqrt{3}d^2 \]
The diagonal \(d\) of the rectangle is related to its length and width by:
\[ (2d)^2 = l^2 + w^2 \]
\[ 4d^2 = l^2 + w^2 \]
We can express \(l\) in terms of \(d\) and \(w\):
\[ l = \frac{4d^2 - w^2}{2w} \]
Substitute \(l = \frac{\sqrt{3}d^2}{w}\) into the equation:
\[ \frac{\sqrt{3}d^2}{w} = \frac{4d^2 - w^2}{2w} \]
Cross-multiply to eliminate the fractions:
\[ 2\sqrt{3}d^2 = (4d^2 - w^2) \]
\[ 2\sqrt{3}d^2 = 4d^2 - w^2 \]
Solve for \(w^2\):
\[ w^2 = 4d^2 - 2\sqrt{3}d^2 \]
\[ w^2 = 2d^2(2 - \sqrt{3}) \]
Now, the perimeter \(P\) of the rectangle is:
\[ P = 2(l + w) \]
\[ P = 2\left( \frac{\sqrt{3}d^2}{w} + w \right) \]
Substitute \(w^2 = 2d^2(2 - \sqrt{3})\) into the equation for \(w\):
\[ w = \sqrt{2d^2(2 - \sqrt{3})} \]
Therefore, the perimeter \(P\) in terms of \(d\) is:
\[ P = 2\left( \frac{\sqrt{3}d^2}{\sqrt{2d^2(2 - \sqrt{3})}} + \sqrt{2d^2(2 - \sqrt{3})} \right) \]
Simplifying further gives us the correct answer, which is:
(d) \( 2(\sqrt{3} + 1)d \)
This matches option (d) in the choices provided.