(Take $\pi=3$ )
$$
\begin{array}{l}
O I=2 \mathrm{~cm}, O J=4 \mathrm{~cm} \\
\cos \angle I O J=\frac{O I}{O J}=\frac{1}{2} \\
\angle I O J=60^{\circ} \\
S_{B I J}=S_{\text {sector } O B J}-S_{\triangle O I J} \\
=\left(\frac{1}{2} \cdot 4^{2} \cdot \frac{\pi}{3}-\frac{1}{2} \cdot 2 \cdot 4 \sin 60^{\circ}\right) \mathrm{cm}^{2} \\
=\left(\frac{8 \pi}{3}-2 \sqrt{3}\right) \mathrm{cm}^{2} \\
\text { Shaded area }=S_{B I J}-S_{B I K} \\
=\left(\frac{8 \pi}{3}-2 \sqrt{3}-\frac{1}{4} \pi \cdot 2^{2}\right) \mathrm{cm}^{2} \\
=\left(\frac{5 \pi}{3}-2 \sqrt{3}\right) \mathrm{cm}^{2} \\
T=5-2 \sqrt{3} \\
\end{array}
$$
G3.4 In Figure 2, the sector $O A B$ has radius $4 \mathrm{~cm}$ and $\angle A O B$ is a right angle. Let the semi-circle with diameter $O B$ be centred at $I$ with $I J / / O A$, and $I J$ intersects the semi-circle at $K$. If the area of the shaded region is $T \mathrm{~cm}^{2}$, find the value of $T$.