Geometry problem - area calculation

Note $\angle OCB =90^{\circ}$ that $\Delta AOC$ is an right-angled triangle and $\overline{CA} = 12$ by Pythagoras' theorem.

area of $\Delta AOC = \dfrac{12 \times 5}{2} = 30$ . Similarly, $\Delta AOB =30$ .

$\angle AOC = \angle AOB = \arctan{\dfrac{12}{5}}$ .

Required area $=$ area of $\Delta AOC$ + area of $\Delta AOB$ - area of sector $BOC$

$= 30 + 30 - \pi \times 5^2 \times \dfrac{2\arctan{\frac{12}{5}}}{360^{\circ}}$

$\approx 30.6.$