In the diagram,  AC represents the diameter of a circle with a radius of $5 \mathrm{cm}$

If $\mathrm{AB}=\mathrm{BC}$ and $CD=8 \mathrm{cm}$, then what is the area of the shaded region rounded off to the nearest whole number?

Given :  $\mathrm{AC}$ is the diameter of a circle with a radius $5 \mathrm{cm}$, $\mathrm{AB}=\mathrm{BC} \mathrm{and} \mathrm{CD}=8 \mathrm{cm}$

$\mathrm{As} \mathrm{AC} \mathrm{is} \mathrm{the} \mathrm{diameter} \mathrm{so},$

$\mathrm{AC} =10 \mathrm{cm} (2\times \mathrm{radius})$

We know that the Area of a circle $=\pi {\mathrm{r}}^2$ square units

Area of the circle $=\pi (5{)}^2=25\pi {\mathrm{cm}}^2$

We know that, the angle subtended by the diameter of a circle at any point on the circle is a right angle.

Hence, $∆\mathrm{ABC}$ is a right-angled triangle with $\angle \mathrm{B}=90°$

$\therefore {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2$ (Using Pythagoras theorem)

$\Rightarrow 100=2 {\mathrm{AB}}^2$             (Given AB = AC)    

$\Rightarrow {\mathrm{AB}}^2=50$                   

$\Rightarrow \mathrm{AB}=\sqrt{50}$                 (By taking square roots on both sides)

$\Rightarrow AB=\sqrt{2\times 5^2}$

$\therefore \mathrm{AB}=\mathrm{BC}=5\sqrt{2}$

$\Rightarrow \mathrm{AB}=5\sqrt{2}$

We know that, Area of a triangle = $\frac{1}{2}\times \mathrm{base} \times \mathrm{height}$

Now, the Area of $∆\mathrm{ABC}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}$      

$\Rightarrow \frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25{ \mathrm{cm}}^2$       ...(i)

Similarly,

$∆\mathrm{ADC}$ is a right-angled triangle with $\angle \mathrm{D}=90°$

$\therefore {\text{AC}}^2={\text{AD}}^2+{\text{DC}}^2$

$\Rightarrow {10}^2={\text{AD}}^2+8^2$

$\Rightarrow 100={\text{AD}}^2+64$

$\Rightarrow {\text{AD}}^2=36$

$\Rightarrow \mathrm{AD}=\sqrt{36}$                 (By taking square roots on both sides)

$\Rightarrow \text{AD}=6$

Now, the Area of $∆\mathrm{ADC}=\frac{1}{2}\times \mathrm{AD}\times \mathrm{DC}$

$\Rightarrow \frac{1}{2}\times 6\times 8=24{ \mathrm{cm}}^2$      ...(ii)

Now, the Area of the given circle is $={\mathrm{\pi r}}^2$

$=\pi \times 5^2$

$=25\mathrm{\pi }$      ...(iii)

$\therefore$ Area of shaded region = Area of the circle $-$ (Area of $∆\mathrm{ABC} +$ Area of  $∆\mathrm{ADC}$)

 $=25\pi -25-24$       {from (i), (ii), and (iii)}

$\begin{array}{c}=25\pi -49\\ =29.5{ \mathrm{cm}}^2\end{array}$

The nearest whole number is 30.

Hence, the area of the shaded region to the nearest whole number is $30{ \mathrm{cm}}^2$