The area of the region \ ( x, y ) : 0 ≤ x ≤ (9)/(4), 0 ≤ y ≤ 1, x ≥...

The region is a quadrilateral $ABCD$

Now, area of the quadrilateral $ABCD =$area of triangle $ABD +$area of the triangle $BCD$

So, area = \dfrac{1}{2}\left|\begin{cases}\dfrac{3}{2} & \dfrac{1}{2} & 1 \\ 2 & 0 & 1 \\ \dfrac{9}{4} & \dfrac{3}{4} & 1\end{cases}\right| + \dfrac{1}{2} \times \left(\dfrac{9}{4} - 2\right) \times \dfrac{3}{4}

$= \dfrac{1}{2}\left\{- 2\left(\dfrac{1}{2} - \dfrac{3}{4}\right) - \left(\dfrac{9}{8} - \dfrac{9}{8}\right)\right\} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4}$

$= \dfrac{1}{4} + \dfrac{3}{32} = \dfrac{11}{32}$