7. The circle inscribed in triangle $A B C$ touches side $B C$ at point $D$, and $B D=5$. Find the radius of this circle and the area of triangle $A B C$, if angle $A$ is $60^{\circ}$ and the radius of the circumscribed circle around triangle $A B C$ is $7 / \sqrt{3}$.
# Solution:

Since $2 R_{\text {on }}=B C / \sin \angle A$, then

$B C=2 R_{\text {on }} \sin \angle A=7$. Then $B D=a=5$, $D C=b=2$. If $F$ and $E$ are the points of tangency of the circle with sides $B A$ and $A C$ respectively, then $B F=a=5, E C=b=2$, and $F A=A E=x$. By the cosine rule we have:

$B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A$. Since $\angle A=60^{\circ}$, then

$49=(x+5)^{2}+(x+2)^{2}-(x+5)(x+2)$, and

$x^{2}+7 x-30=0, x=3$. Then the area

![](https://cdn.mathpix.com/cropped/2024_05_06_450936e33dcef50d2d99g-08.jpg?height=626&width=720&top_left_y=1189&top_left_x=1225)

$S_{A B C}=0.5(5+x)(2+x) \sin \angle A=10 \sqrt{3}$, perimeter

$P_{A B C}=2(a+b+x)=2 \cdot 10$. The radius of the inscribed circle $r_{\epsilon n}=2 S_{A B C} / P_{A B C}=\sqrt{3}$.

Answer: $S_{A B C}=10 \sqrt{3}, r_{b n}=\sqrt{3}$.