check the picture below.
now, bear in mind that, we could simply get the area of the whole circle with that radius, and tease out a quarter, because the segment is just using up a quarter of the circle, because the angle made is 90°, and then subtract the triangle from that sector, and what's leftover is the segment.
[tex]\bf \stackrel{\textit{area of the circle}}{A=\pi r^2}\implies A=\pi (3\sqrt{2})^2\implies A=\pi 3^2\sqrt{2^2}\implies A=18\pi
\\\\\\
\textit{now one-quart of that is }\cfrac{18\pi }{4}\implies \cfrac{9\pi }{2}\impliedby \textit{sector's area}[/tex]
[tex]\bf \stackrel{\textit{area of the triangle}}{A=\cfrac{1}{2}bh}\implies A=\cfrac{(3\sqrt{2})(3\sqrt{2})}{2}\implies A=\cfrac{3^2\sqrt{2^2}}{2}\\\\\\A=\cfrac{18}{2}\implies
A=9\impliedby \textit{area of that triangle}\\\\
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\stackrel{sector's area}{\cfrac{9\pi }{2}}~~-~~\stackrel{\textit{area of the triangle}}{9}\implies \cfrac{9\pi -18}{2}\impliedby \textit{segment's area}[/tex]
or you can always just use the area of a segment, with the radius and angle given.
[tex]\bf \textit{area of a segment of a circle}\\\\
A=\cfrac{r^2}{2}\left[\cfrac{\theta \pi }{180}-sin(\theta ) \right]
\\\\\\
\begin{cases}
r=3\sqrt{2}\\
\theta =90
\end{cases}\implies A=\cfrac{(3\sqrt{2})^2}{2}\left[\cfrac{90 \pi }{180}-sin(90^o ) \right][/tex]
