To solve the problem, we first find the height and width of the inscribed triangle.

To calculate the measure of $\mathrm{\angle OBA}$, we draw a perpendicular line from the center to chord $\mathrm{AB}$, which bisects the chord.



Height of triangle $\mathrm{△OBA}$ $=$ radius $⨯$ $\sin \left(\mathrm{\angle OBA}\right)$

$\frac{\mathrm{AB}}{2}$ or half base width of the triangle $\mathrm{△OBA}$ $=$ radius $⨯$ $\cos \left(\mathrm{\angle OBA}\right)$

The measure of $\mathrm{\angle OBA}$ is calculated as $180-\left(90+\frac{90}{2}\right)$ degrees. One angle is $90$ degrees because the perpendicular line from the center to chord $\mathrm{AB}$ also bisects $\mathrm{\angle AOB}$.

The height of the triangle is $2\sin \left(180-\left(90+\frac{90}{2}\right)\right)=\sqrt{2}$.

Half the triangle width is $2\cos \left(180-\left(90+\frac{90}{2}\right)\right)=\sqrt{2}$.

Therefore, the full width of the triangle is $2⨯\sqrt{2}=2\sqrt{2}$.

Now, we calculate the area of triangle $\mathrm{△OBA}$, which is given by Area $=$ $\frac{1}{2}$ $⨯$ base $⨯$ height.

Substitute the values:

Area of triangle $\mathrm{△OBA}$ $=$ $\frac{1}{2}⨯\left(2\sqrt{2}\right)⨯\sqrt{2}=2$

The area of the circular sector subtended by the minor angle $\mathrm{\angle AOB}$ is $\frac{90}{360}⨯\left(\pi 2^2\right)=\pi$.

The shaded area is obtained by subtracting the area of triangle $\mathrm{△OBA}$ from the area of the circular sector subtended by the minor angle $\mathrm{\angle AOB}$.

Area of the shaded region $=$ $\pi -2=1.14$ (rounded to the nearest hundredth)