Answer:
[tex](2 \pi-4)\ units^2[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The area of the region restricted by the circle and outside of the square is equal to the area of circle minus the area of square
step 1
Find the area of square
The area of square is
[tex]A=b^{2}[/tex]
where
b is the length side of square
we have
[tex]b=2\ in[/tex]
substitute
[tex]A=2^{2}=4\ units^2[/tex]
step 2
Find the area of the circle
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
where
r is the radius of the circle
In this problem
The diameter of the circle is equal to the diagonal of the square
Find the diagonal of the square
Applying Pythagorean Theorem
[tex]D=\sqrt{b^2+b^2}[/tex]
[tex]b=2\ in[/tex]
substitute
[tex]D=\sqrt{2^2+2^2}[/tex]
[tex]D=\sqrt{8}\ units[/tex]
simplify
[tex]D=2\sqrt{2}\ units[/tex]
Find the radius of the circle
[tex]r=\frac{2\sqrt{2}}{2}=\sqrt{2}\ units[/tex] ---> the radius is half the diameter
Find the area of the circle
[tex]A=\pi r^{2}[/tex] ----> [tex]A=\pi(\sqrt{2})^{2}=2 \pi\ units^2[/tex]
step 3
Find the area of the region restricted by the circle and outside of the square
Subtract the area of square from the area of the circle
[tex](2 \pi-4)\ units^2[/tex]
