Question

In the given figure, $O$ is the centre of the circle with radius $28\text{cm}$. If $AB$ be the diameter of semicircle, then area of the shaded region is $[\text{use}\pi =\frac{22}{7}]$

• A. $616{\text{cm}}^2$
• B. $308{\text{cm}}^2$
• C. $392{\text{cm}}^2$
• D. $794{\text{cm}}^2$

Answer 1

The correct option is C $392{\text{cm}}^2$

Radius of circle, $r=28\text{cm}$
Area of the minor segment formed by chord $AB=$ Area of the minor sector $OAB–$ Area of $\Delta OAB$
Area of minor sector $OAB=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times (28{)}^2$

$(\theta ={90}^{\circ })$

$=\frac{\pi }{4}\times 28\times 28$

$=\frac{22\times 28\times 28}{7\times 4}$
$=616{\text{cm}}^2$

Area of $\Delta OAB=\frac{1}{2}\times OA\times OB$
$=\frac{1}{2}\times 28\times 28$
$=392{\text{cm}}^2$

$\therefore$ Area of minor segment formed by the chord $AB=616–392=224{\text{cm}}^2$
Now, area of the shaded region
= Area of semicircle with diameter $AB–$ Area of minor segment formed by the chord $AB$

In $\Delta AOB$, by Pythagoras theorem,
$A{B}^2=A{O}^2+O{B}^2$
$\Rightarrow A{B}^2=(28{)}^2+(28{)}^2$
$\Rightarrow A{B}^2=2\times (28{)}^2$
$\Rightarrow AB=28\sqrt{2}$
$\Rightarrow \frac{AB}{2}=\frac{28\sqrt{2}}{2}=14\sqrt{2}\text{cm}$ ... (ii)
Substituting (ii) in (i), we get
Area of the shaded region $=[\frac{1}{2}\times \frac{22}{7}\times (14\sqrt{2}{)}^2]-224$
$=[\frac{1}{2}\times \frac{22}{7}\times (14\sqrt{2})\times (14\sqrt{2}\left)\right]-224$
$=616-224$
$=392{\text{cm}}^2$

Hence, the correct answer is option c.