ABC26GN4335 · Area
Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
If the area of a rectangle is $\sqrt{3} d^{2}$, where $2 d$ is the length of its diagonal, then its perimeter is equal to
(a)$4 \sqrt{3} d$
(b)$2 \sqrt{3} d$
(c)$4(\sqrt{3}+1) d$
(d)$2(\sqrt{3}+1) d$
Answer
Explanation
$\sqrt{l^{2}+b^{2}}=2 d \Rightarrow l^{2}+b^{2}=4 d^{2}$. Also, $l b=\sqrt{3} d^{2}$. $$\begin{aligned} & (l+b)^{2}=\left(l^{2}+b^{2}\right)+2 l b=4 d^{2}+2 \sqrt{3} d^{2} \\ & \Rightarrow(l+b)=\sqrt{(4+2 \sqrt{3}) d^{2}}=\sqrt{\left[(1)^{2}+(\sqrt{3})^{2}+2 \sqrt{3}\right] d^{2}} \\ & =\sqrt{(\sqrt{3}+1)^{2} d^{2}}=(\sqrt{3}+1) d \\ & \therefore \quad \text { Perimeter }=2(l+b)=2(\sqrt{3}+1) d \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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