Question

PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS = 12 cm, find the perimeter and area of the shaded region.

Answer 1

Perimeter (circumference of the circle) = $2\mathrm{\pi r}$
We know:
Perimeter of a semicircular arc = $\mathrm{\pi r}$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = $\mathrm{\pi r}=6\mathrm{\pi }\mathrm{cm}$

For the arc QES, radius is 4 cm.
​∴ Circumference of the semicircle QES = $\mathrm{\pi r}=4\mathrm{\pi }\mathrm{cm}$

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = $\mathrm{\pi r}=2\mathrm{\pi }\mathrm{cm}$

Now,
Perimeter of the shaded region = $6\mathrm{\pi }+4\mathrm{\pi }+2\mathrm{\pi }$
$=12\mathrm{\pi cm}$
$$\begin{gathered} =12\times 3.14 \\ =37.68\mathrm{cm} \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{2}\times 3.14\times 2\times 2 \\ =6.28{\mathrm{cm}}^2 \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{2}\times 3.14\times 6\times 6 \\ =56.52{\mathrm{cm}}^2 \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{2}\times 3.14\times 4\times 4 \\ =25.12{\mathrm{cm}}^2 \end{gathered}$$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES
$=6.28+56.52-25.12=37.68{\mathrm{cm}}^2$