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1. The length of a rectangle is three times of its width. If the length of the diagonal is $8 \sqrt{10}$ cm, then the perimeter of the rectangle is :

A. $15 \sqrt{10}$ cm

B. $16 \sqrt{10}$ cm

C. $24 \sqrt{10}$ cm

D. 64 cm

Solution:

Let breadth = x cm
Then, length = 3x cm

\begin{aligned} \Rightarrow x^{2} + {(3 x)}^{2} = {(8 \sqrt{10})}^{2} \\ \Rightarrow 10 x^{2} = 640 \\ \Rightarrow x^{2} = 64 \\ \Rightarrow x = 8 \end{aligned}

So, length = 24 cm and breadth = 8 cm
∴ Perimeter = [2(24 + 8)] cm = 64 cm

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