The diagonal of a rectangle is √(41) cm and its area is 20 sq. cm. The perimeter of the rectangle must be · ABC26GN4334
Subject: General Aptitude · Chapter: Area · Exam: Hotel Management 2002
The diagonal of a rectangle is $\sqrt{41} \mathrm{cm}$ and its area is 20 sq. cm. The perimeter of the rectangle must be
(a)9 cm
(b)18 cm
(c)20 cm
(d)41 cm
Answer
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Explanation
$\sqrt{l^{2}+b^{2}}=\sqrt{41}$ or $l^{2}+b^{2}=41$. Also, $l b=20$. $$(l+b)^{2}=\left(l^{2}+b^{2}\right)+2 l b=41+40=81 \Rightarrow(l+b)=9 .$$ ∴ Perimeter $=2(l+b)=18 \mathrm{cm}$.
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