PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal.

Solution:

Perimeter (circumference of the circle) $=2 \pi \mathrm{r}$

We know:

Perimeter of a semicircular $\operatorname{arc}=\pi \mathrm{r}$

Now,
For the arc PTS, radius is 6 cm.

$\therefore$ Circumference of the semicircle PTS $=\pi \mathrm{r}=6 \pi \mathrm{cm}$

For the arc QES, radius is 4 cm.

$\therefore$ Circumference of the semicircle $Q E S=\pi r=4 \pi \mathrm{cm}$

For the arc PBQ, radius is 2 cm.

$\therefore$ Circumference of the semicircle $\mathrm{PBQ}=\pi \mathrm{r}=2 \pi \mathrm{cm}$

Now,

Perimeter of the shaded region $=6 \pi+4 \pi+2 \pi$

$=12 \pi \mathrm{cm}$

$=12 \times 3.14$

$=37.68 \mathrm{~cm}$

Area of the semicircle $P B Q=\frac{1}{2} \pi r^{2}$

$=\frac{1}{2} \times 3.14 \times 6 \times 6$

$=56.52 \mathrm{~cm}^{2}$

Area of the semicircle $\mathrm{QES}=\frac{1}{2} \pi \mathrm{r}^{2}$

$=\frac{1}{2} \times 3.14 \times 4 \times 4$

$=25.12 \mathrm{~cm}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS -">− Area of the semicircle QES

$=6.28+56.52-25.12=37.68 \mathrm{~cm}^{2}$