Question

In figure 6.58, seg PQ is a diameter of semicircle PNQ. The centre of arc PMQ is O. OP = OQ = 10 cm and m $\angle$POQ = 60$°$. Find the area of the shaded portion. (Given $\mathrm{\pi }=3.14,\sqrt{3}=1.73$)

Answer 1

Area of segment PMQ = r² $\left[\frac{\mathrm{\pi \theta }}{360}-\frac{\mathrm{sin\theta }}{2}\right]$

=${10}^2\left[\frac{3.14\times 60}{360}-\frac{\sin 60°}{2}\right]$

= $100\left[0.52-\frac{\sqrt{3}}{4}\right]$

= $100\left[0.52-0.43\right]$
= 9 cm²
Now,
In $∆$OPQ,
OP = OQ = 10 cm (Radii of the circle)
We know that $∆$OPQ is an isosceles triangle.
Now, we will draw a perpendicular from vertex A that will bisect the opposite side BF at M.

We have:
$\angle$O = 60$°$
∴ $\angle$AOQ = $\frac{1}{2}$$\angle$O = $\frac{1}{2}$× 60$°$ = 30$°$
Sin $\angle$AOQ = $\frac{\mathrm{AQ}}{\mathrm{OQ}}$

$\Rightarrow$sin 30$°$ = $\frac{\mathrm{AQ}}{10}$

$\Rightarrow$AQ = 10 × $\frac{1}{2}$ cm = 5 cm

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

$$\begin{gathered} =\frac{3.14\times 5\times 5}{2} \\ =39.25{\mathrm{cm}}^2 \end{gathered}$$

Now, we have:
Area of the shaded region = Area of the semicircle $-$ Area of the segment
= 39.25 $-$ 9
= 30.25 cm²