In the adjoining figure,$PS$ is the diameter of a circle and $PS = 12$. $PQ = QR = RS$. Semicircles are drawn with diameters $\overline{PQ}$ and $\overline{QS}$. Find the perimeter and the area of the shaded region. $(\pi = 3.14)$

(N/A) Here,$PS = 12 \text{ cm}$ and $PQ = QR = RS$.
$\therefore PQ = QR = RS = \frac{12}{3} = 4 \text{ cm}$.
The radii of semicircles with diameters $\overline{PS}$,$\overline{QS}$,and $\overline{PQ}$ are:
$r_1 = \frac{PS}{2} = \frac{12}{2} = 6 \text{ cm}$,
$r_2 = \frac{QS}{2} = \frac{4+4}{2} = 4 \text{ cm}$,and
$r_3 = \frac{PQ}{2} = \frac{4}{2} = 2 \text{ cm}$ respectively.
Perimeter of the shaded region:
$= \text{Sum of the lengths of all the three semicircular arcs}$
$= \pi r_1 + \pi r_2 + \pi r_3$
$= \pi(r_1 + r_2 + r_3)$
$= 3.14(6 + 4 + 2)$
$= 37.68 \text{ cm}$.
Area of the shaded region:
$= \text{Area of the semicircle with radius } r_1 + \text{Area of the semicircle with radius } r_3 - \text{Area of the semicircle with radius } r_2$
$= \frac{1}{2} \pi r_1^2 + \frac{1}{2} \pi r_3^2 - \frac{1}{2} \pi r_2^2$
$= \frac{1}{2} \pi(r_1^2 + r_3^2 - r_2^2)$
$= \frac{1}{2} \times 3.14(6^2 + 2^2 - 4^2)$
$= \frac{1}{2} \times 3.14 \times (36 + 4 - 16)$
$= \frac{1}{2} \times 3.14 \times 24$
$= 37.68 \text{ cm}^2$.
Thus,the perimeter of the shaded region is $37.68 \text{ cm}$ and its area is $37.68 \text{ cm}^2$.