The length of a rectangle is twice its width. If the length of its diagonal is $16 \sqrt{5}$ cm, find its area.

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Solution

Let breadth be $x$ and length be $2 x$
$\sqrt{(4 x^{2} + x^{2})} = \sqrt{5} x \sqrt{5} x = 16 \sqrt{5} x = 16 l e n g t h = 32 b r e a d t h = 16 A r e a = l e n g t h \times b r e a d t h = (16 \times 32) c m^{2} = 512 c m^{2}$

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