7. The inscribed circle of a triangle divides one of its sides into segments equal to 3 and 4. Find the area of the triangle if the radius of the circumscribed circle around it is $7 / \sqrt{3}$.
# Solution:

Let $D$ be the point of tangency of the circle with side $B C$.

By the condition, $B D=a=3, D C=b=4$. Therefore, $B C=a+b=7$. Since $2 R_{\text{on}}=B C / \sin \angle A$, we have $\sin \angle A=B C / 2 R_{\text{on}}=\sqrt{3} / 2$. Thus, $\angle A=60^{\circ}$ or $\angle A=120^{\circ}$. If $F$ and $E$ are the points of tangency of the circle with sides $B A$ and $A C$ respectively, then $B F=a=3, E C=b=4$, and $F A=A E=x$. By the cosine rule, we have:

$$
B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A .
$$

1) If $\angle A=60^{\circ}$, then

$49=(x+3)^{2}+(x+4)^{2}-(x+3)(x+4)$, and

![](https://cdn.mathpix.com/cropped/2024_05_06_450936e33dcef50d2d99g-03.jpg?height=623&width=714&top_left_y=748&top_left_x=1225)

$x^{2}+7 x-36=0, x=\frac{\sqrt{193}-7}{2}$.

Then $S_{A B C}=\frac{1}{2}(3+x)(4+x) \sin \angle A=\frac{\sqrt{3}}{4}\left(\frac{\sqrt{193}-1}{2}\right)\left(\frac{\sqrt{193}+1}{2}\right)=12 \sqrt{3}$.

2) If $\angle A=120^{\circ}$, then $49=(x+3)^{2}+(x+4)^{2}+(x+3)(x+4)$, and $x^{2}+7 x-4=0$, $x=\frac{\sqrt{65}-7}{2}$. Then $\quad S_{A B C}=\frac{1}{2}(3+x)(4+x) \sin \angle A=\frac{\sqrt{3}}{4}\left(\frac{\sqrt{65}-1}{2}\right)\left(\frac{\sqrt{65}+1}{2}\right)=4 \sqrt{3}$.

Answer: $12 \sqrt{3}$ or $4 \sqrt{3}$.