Answer
The area of the shaded region of the given figure, in terms of \[\pi \] is\[\left( \text{12}\text{.5}\pi -24 \right)in{{.}^{2}}\].
Work Step by Step
We have to find the area of the shaded region in the given figure. In the figure, a right angled triangle is within the bigger circle. The area of the shaded region can be found out by subtracting the area of the triangle, with base equal to \[6\text{ inches}\] and the height equal to \[\text{8 inches}\],from the area of the circle.
Compute the diameter of the circle by applying the Pythagoras theorem to the right angled triangle, as follows:
\[\begin{align}
& d=\sqrt{8\text{ i}{{\text{n}}^{2}}+6\text{ i}{{\text{n}}^{2}}} \\
& =\sqrt{100\text{ in}} \\
& =10\text{ in}
\end{align}\]
So, the radius will be half of the diameter, that is \[\frac{10\text{ inches}}{2}=5\text{ inches}\].
According to this formula, the area of the semicircle, with radius equal to \[5\text{ inches}\] will be:
\[\begin{align}
& {{A}_{1}}=\frac{1}{2}\pi {{r}^{2}} \\
& =\frac{1}{2}\pi \times 5\text{ in}{{\text{.}}^{2}} \\
& =12.5\pi \text{ in}{{\text{.}}^{2}}
\end{align}\]
The area of the triangle, with base equal to \[6\text{ inches}\] and the height equal to \[\text{8 inches}\]will be:
\[\begin{align}
& {{A}_{2}}=\frac{1}{2}bh \\
& =\frac{1}{2}\times 6\text{ in}\text{.}\times 8\text{ in}\text{.} \\
& =24\text{ in}{{.}^{2}}
\end{align}\]
The area of the shaded region will be:
\[\begin{align}
& A={{A}_{1}}-{{A}_{2}} \\
& =\left( \text{12}\text{.5}\pi -24 \right)in{{.}^{2}}
\end{align}\]
Hence, the area of the shaded region of the given figure, in terms of \[\pi \] is\[\left( \text{12}\text{.5}\pi -24 \right)in{{.}^{2}}\]
