Question

In the given figure, $PQRS$ is a diameter of a circle of radius $6$ $cm$. The lengths $PQ,QR$ and $RS$ are equal. Semi-circles are drawn on $PQ$ and $QS$ as diameters. If $PS=12$ $cm$, find the perimeter and the area of the shaded region.
[Take $\pi =3.14$]

• A. $38.68$ $cm$, $38.68$ $c{m}^2$
• B. $40.68$ $cm$, $42.68$ $c{m}^2$
• C. $37.68$ $cm$, $37.68$ $c{m}^2$
• D. $45.68$ $cm$, $36.62$ $c{m}^2$

Answer 1

The correct option is C $37.68$ $cm$, $37.68$ $c{m}^2$

Given, $PS=12$ $cm$
As $PQ=QR=RS$
$PQ=QR=RS=\frac{1}{3}PS=\frac{12}{3}=4$ $cm$
$QS=2PQ$
$\Rightarrow QS=2\times 4=8$ $cm$
$\therefore$ area of the shaded region $=$ Area of the semicircle with PS as the diameter $+$ Area of the semicircle with $PQ$ as the diameter $-$ Area of the semicircle with $QS$ as the diameter
$=\frac{1}{2}[3.14\times 6^2+3.14\times 2^2-3.14\times 4^2]$
$=\frac{1}{2}[3.14\times 36+3.14\times 4-3.14\times 16]$
$=\frac{3.14}{2}[36+4-16]$
$=\frac{3.14}{2}\left[24\right]$
$=37.68c{m}^2$
Perimeter of the shaded region $=$ Arc of semicircle with $PS$ as diameter $+$ Arc of semicircle with $QS$ as diameter $+$ Arc of semicircle with $PQ$ as diameter
$=\frac{1}{2}[2\times 3.14\times 6+2\times 3.14\times 4+2\times 3.14\times 2]$
$=3.14[6+4+2]$
$=3.14\left[12\right]$
$=37.68cm$