Question:
The length of a rectangle is thrice its breadth and the length of its diagonal is $8 \sqrt{10} \mathrm{~cm}$. The perimeter of the rectangle is
(a) $15 \sqrt{10} \mathrm{~cm}$
(b) $16 \sqrt{10} \mathrm{~cm}$
(c) $24 \sqrt{10} \mathrm{~cm}$
(d) 64 cm
Solution:
(d) 64 cm
Let the breadth of the rectangle be x cm.
∴ Length of the rectangle = 3x cm
We know:
Diagonal $=\sqrt{(\text { Length })^{2}+(\text { Breadth })^{2}}$
$\Rightarrow 8 \sqrt{10}=\sqrt{x^{2}+(3 x)^{2}}$
$\Rightarrow 8 \sqrt{10}=\sqrt{x^{2}+9 x^{2}}$
$\Rightarrow 8 \sqrt{10}=x \sqrt{10}$
$\Rightarrow x=8$
Now,
Breadth of the rectangle = x = 8 cm
Length of the rectangle = 3x = 24 cm
Perimeter of the rectangle $=2($ Length $+$ Breadth $)=2(8+24)=64 \mathrm{~cm}$