10.105. Find the area of an isosceles triangle (Fig. 10.101) with an angle of $120^{\circ}$, if the radius of the inscribed circle is $\sqrt[4]{12}$ cm.
## Solution.

From $\triangle B D C$, we find $B D=B C \cdot \sin 30^{\circ}=\frac{B C}{2} ; D C=B C \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{2} B C$. The semiperimeter of $\triangle A B C \quad p=\frac{1}{2}(x+x+\sqrt{3} x)=x\left(1+\frac{\sqrt{3}}{2}\right)$.

Area of $\triangle A B C: S=p r=x\left(1+\frac{\sqrt{3}}{2}\right) \sqrt[4]{12}$.

On the other hand, $S=\frac{1}{2} A C \cdot B D=\frac{1}{2} \sqrt{3} x \cdot \frac{x}{2}=\frac{\sqrt{3}}{4} x^{2}$. Solving the equation $\frac{\sqrt{3}}{4} x^{2}=x\left(1+\frac{\sqrt{3}}{2}\right) \sqrt[4]{12}$. We get $x=\frac{2(2+\sqrt{3}) \sqrt[4]{12}}{\sqrt{3}}$. Then the area of $\triangle A B C \quad S=\frac{\sqrt{3}}{4}\left(\frac{2(2+\sqrt{3}) \sqrt[4]{12}}{\sqrt{3}}\right)^{2}=2(7+4 \sqrt{3})\left(\mathrm{cm}^{2}\right)$.

Answer: $\quad 2(7+4 \sqrt{3})\left(\mathrm{cm}^{2}\right)$.