8. In Rt $\triangle A B C$, $\angle A=30^{\circ}$, the angle bisector of $\angle A$ is $1 \mathrm{~cm}$. Then, the area of $\triangle A B C$ is $\qquad$
8. $\frac{3+2 \sqrt{3}}{24}$.

As shown in Figure 5, $\frac{A C}{A B}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$,
thus, $\frac{C D}{D B}=\frac{A C}{A B}=\frac{\sqrt{3}}{2}$.
Let $C D=\sqrt{3} x, D B=2 x$, then
$$
B C=(2+\sqrt{3}) x, A C=\sqrt{3} B C=\sqrt{3}(2+\sqrt{3}) x \text {. }
$$

Applying the Pythagorean theorem, we get
$3 x^{2}+3(\sqrt{3}+2)^{2} x^{2}=1$, solve for $x^{2}$.