SOLUTION
We want to solve the question below
The figure consists of a semi-circle and a triangle. So the area of the figure becomes
Area of semi-circle + area of triangle
The semi-circle has a diameter of 8 in. So the radius becomes
[tex]r=\frac{diameter}{2}=\frac{8}{2}=4in[/tex]
Area of the semi-circle is given as
[tex]\begin{gathered} \frac{1}{2}\times\pi r^2 \\ \frac{1}{2}\times3.14\times4^2 \\ \frac{1}{2}\times3.14\times16 \\ =25.12\text{ in}^2 \end{gathered}[/tex]
Area of the triangle is
[tex]\begin{gathered} \frac{1}{2}\times base\times height \\ \frac{1}{2}\times9\times8 \\ 9\times4 \\ =36\text{ in}^2 \end{gathered}[/tex]
So Area of the figure becomes
[tex]\begin{gathered} 25.12+36 \\ =61.12\text{ in}^2 \end{gathered}[/tex]
Hence the answer is option C
