Three squares $ P, Q $ and $ R $ are such that the perimeter of $ \mathrm{P} $ is $ \frac{2}{3} $ the perimeter of $ \mathrm{Q} $, and the perimeter of $ \mathrm{Q} $ is $ \frac{2}{3} $ the perimeter of $ \mathrm{R} $. If the area of $ \mathrm{P} $ is $ 16 \mathrm{sq} $. units, what is the area of $ \mathrm{R} $ ?

(1) 9 sq. units
(2) 81 sq. units
(3) 64 sq. units
(4) 36 sq. units

Given:

The perimeter of \( \mathrm{P} \) is \( \frac{2}{3} \) the perimeter of \( \mathrm{Q} \), and the perimeter of \( \mathrm{Q} \) is \( \frac{2}{3} \) the perimeter of \( \mathrm{R} \).

The area of \( \mathrm{P} \) is \( 16 \mathrm{sq} \). units.
To do:

We have to find the area of $R$.

Solution:

Let the side of square R be $x$.

This implies the perimeter of square R$=4x$.

The perimeter of square Q$=\frac{2}{3}\times4x$

$=\frac{8x}{3}$

The perimeter of square P$=\frac{2}{3}\times\frac{8x}{3}=\frac{16x}{9}$

This implies,

Side of square P$=\frac{\frac{16x}{9}}{4}=\frac{4x}{9}$

Area of square P$=(\frac{4x}{9})^2=\frac{16x^2}{81}$

Therefore,

$\frac{16x^2}{81}=16$

$x^2=\frac{16\times81}{16}$

$x^2=81$

$x=\sqrt{81}$

$x=9$

Side of square R$=9$ units.

Area of square R$=(9)^2=81$ sq. units.

The correct answer is (2) 81 sq. units.

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Updated on: 10-Oct-2022

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