Answer:
the answer is the first option
[tex](16\pi\ -32)\ in^{2}[/tex]
Step-by-step explanation:
we know that
The area of the shaded portion of the circle is equal to the area of a quarter circle minus the area of the right triangle
Step 1
Find the area of a quarter circle
[tex]A=\frac{1}{4}\pi r^{2}[/tex]
we have
[tex]r=8\ in[/tex]
substitute
[tex]A=\frac{1}{4}\pi (8)^{2}=16\pi\ in^{2}[/tex]
Step 2
Find the area of the right triangle
we know that
the area of the triangle is equal to
[tex]A=\frac{1}{2}bh[/tex]
in this problem we have
[tex]b=h=8\ in[/tex]
substitute
[tex]A=\frac{1}{2}(8)^{2}=32\ in^{2}[/tex]
Step 3
Find the area of the shaded portion
Remember that
The area of the shaded portion of the circle is equal to the area of a quarter circle minus the area of the right triangle
[tex]A=(16\pi\ -32)\ in^{2}[/tex]