In the adjotning flgure, $PS$ is diemeter of a circle and $PS$ $=12$. $P Q=Q R=R S$ Semicircles are drawn with dinmeter $\overline{\text { PQ }}$ and $\overline{QS}$. Find the perimeter and the area Find the perimeter and the arce of the shaded region. $(\pi=3.14)$

Here, $PS =12 cm$ and $PQ = QR = RS$

$\therefore PO = gR = RS =\frac{12}{3}=4 cm$

The radii of semicircles with diameters $\overline{ PS }, \overline{ QS }$ and $\overline{ PQ }$ are

$r_{1}=\frac{ PS }{2}=\frac{12}{2}=6 cm$

$r_{2}=\frac{g S}{2}=\frac{4+4}{2}=4 cm$ and

$r_{3}=\frac{ PQ }{2}=\frac{4}{2}=2 cm$ respectively.

Perimeter of the shaded region

$=$ Sum of the lengths of all the three semicircular arcs

$=\pi r_{1}+\pi r_{2}+\pi r_{3}$

$=\pi\left(r_{1}+r_{2}+r_{3}\right)$

$=3.14(6+4+2)$

$=37.68 cm$

Area of the shaded region

$=$ Area of the semicircle with radius $r_{1}$

$+$ Area of the semicircle with radius $r_{3}$

$-$ Area of the semicircle with radius $r_{2}$

$=\frac{1}{2} \pi r_{1}^{2}+\frac{1}{2} \pi r_{3}^{2}-\frac{1}{2} \pi r_{2}^{2}$

$=\frac{1}{2} \pi\left(r_{1}^{2}+r_{3}^{2}-r_{2}^{2}\right)$

$=\frac{1}{2} \times 3.14\left(6^{2}+2^{2}-4^{2}\right)$

$=\frac{1}{2} \times 3.14 \times 24$

$=37.68 cm ^{2}$

Thus, the perimeter of the shaded region is $37.68 cm$ and its area is $37.68 cm ^{2}$