Question

In figure AB is the diameter of the circle, $ \mathrm{AC}=6 \mathrm{cm}$ and $ \mathrm{BC}=8 \mathrm{cm}$. Find the area of the shaded region (Use $ \mathrm{\pi }=3.14$).

Answer 1

Step 1: Find the area of the triangle ABC.

Given: $\mathrm{AC}=6\mathrm{cm}$ and $\mathrm{BC}=8\mathrm{cm}$

Since the angle on the semicircle is right-angled, therefore angle $\angle \mathrm{ACB}=90°.$

Area of the triangle $=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BC}$

$$\begin{gathered} =\frac{1}{2}\times 6\times 8 \\ =24{\mathrm{cm}}^2 \end{gathered}$$

Step 2: Find the area of the circle.

Since ABC is the right-angled triangle.

$$\begin{gathered} \mathrm{AB}=\sqrt{\mathrm{AC}+\mathrm{BC}}(\mathrm{Pythagoras}\mathrm{Theorem}) \\ \mathrm{AB}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2} \\ \mathrm{AB}=\sqrt{36+64} \\ \mathrm{AB}=\sqrt{100}=10\mathrm{cm} \end{gathered}$$

AB is the diameter of the circle.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}={\mathrm{\pi r}}^2 \\ =\frac{22}{7}\times {\left(5\right)}^2(\mathrm{radius}=\frac{\mathrm{diameter}}{2}) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=78.57{\mathrm{cm}}^2 \end{gathered}$$

Step 3: Area of the shaded region.

Area of shaded region $=$Area of the circle$-$Area of the triangle

$$\begin{gathered} =78.57-24 \\ =54.57{\mathrm{cm}}^2 \end{gathered}$$

Hence the area of the shaded region is $54.57{\mathrm{cm}}^2$.