The length of a rectangle is thrice its breadth and the length of its diagonal is
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 = = 8 cm
Length of the rectangle = 3x = 24 cm

Perimeter of the rectangle $=2($ Length $+$ Breadth $)=2(8+24)=64 \mathrm{~cm}$

 

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