The diagonal of a rectangle is $$\sqrt {41} $$ cm and its area is 20 sq. cm. The

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Question

The diagonal of a rectangle is $$\sqrt {41} $$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be :

Options:

A . 9 cm

B . 18 cm

C . 20 cm

D . 41 cm

Answer: Option B
$$\eqalign{
& \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr
& \Rightarrow {l^2} + {b^2} = 41 \cr
& Also,\,\,lb = 20 \cr
& {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2lb \cr
& {\left( {l + b} \right)^2} = 41 + 40 \cr
& {\left( {l + b} \right)^2} = 81 \cr
& \left( {l + b} \right) = 9 \cr
& \therefore {\text{ Perimeter}} = 2\left( {l + b} \right) = 18\,cm \cr} $$

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