check the picture below.
notice, the figure is really just a triangle on top of a semi-circle.
now, the triangle has a base and height of 3 each, and the semi-circle has a diameter of 3√(2), so its radius is half that.
[tex]\bf \begin{array}{llll}
\textit{area of a circle}\\\\
A=\pi r^2
\end{array}\qquad
\begin{array}{llll}
\textit{area of a semi-circle}\\\\
A=\cfrac{\pi r^2}{2}
\end{array}\qquad
\begin{array}{llll}
\textit{area of a triangle}\\\\
A=\cfrac{1}{2}bh
\end{array}\\\\
-------------------------------\\\\[/tex]
[tex]\bf d=3\sqrt{2}\qquad r=\cfrac{3\sqrt{2}}{2}\impliedby \textit{radius of the semi-circle}\\\\
-------------------------------\\\\
\stackrel{\textit{area of triangle}}{\cfrac{3\cdot 3}{2}}+\stackrel{\textit{area of semi-circle}}{\cfrac{\pi \left( \frac{3\sqrt{2}}{2} \right)^2}{2}}\implies \cfrac{9}{2}+\cfrac{\frac{\pi \cdot 3^2\cdot 2}{2}}{2}\implies \cfrac{9}{2}+\cfrac{9\pi }{2}
\\\\\\
\cfrac{9+9\pi }{2}[/tex]
