Then, the area of 1/4 of the circle is:
[tex]\begin{gathered} A=\text{ }\frac{\theta}{360}\text{ x }\pi r^2 \\ A=\text{ }\frac{90}{360}\text{ x }\pi r^2 \\ A\text{ = }\frac{1}{4}\pi\text{ 6}^2 \\ A=\text{ 9}\pi \\ \\ \end{gathered}[/tex]The area of the triangle is:
[tex]\begin{gathered} A=\text{ }\frac{b\text{ x h }}{2} \\ A\text{ = }\frac{6\text{ x 6}}{2} \\ A=\text{ 18in}^2 \end{gathered}[/tex]The area of the shaded region is the area of 1/4 of the circle minus the area of the triangle:
[tex]\begin{gathered} A\text{ = 9}\pi\text{ - 18 in}^2 \\ A=\text{ 28.27in}^2\text{ - 18in}^2 \\ A=\text{ 10.27in}^2 \end{gathered}[/tex]