8.1. (12 points) In triangle $ABC$, the bisector $BL$ is drawn. Find the area of the triangle if it is known that $AL=2, BL=3\sqrt{10}$, and $CL=3$.
Answer: $\frac{15 \sqrt{15}}{4} \approx 14.52$.

Solution. By the property of the angle bisector of a triangle $A B: B C=A L: L C=2: 3$. Therefore, if $A B=2 x$, then $B C=3 x$. Then, since $B L^{2}=A B \cdot B C-A L \cdot L C$, we get $90=6 x^{2}-6$, from which $x=4$. Thus, triangle $A B C$ has sides 8, 12, and 5. From this, using Heron's formula, the area of triangle $A B C$ is

$$
S=\sqrt{p \cdot(p-A B) \cdot(p-B C) \cdot(p-A C)}=\sqrt{\frac{25}{2} \cdot \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{15}{2}}=\frac{15 \sqrt{15}}{4} \approx 14.52
$$