10.405 A circle is inscribed in an equilateral triangle with side $a$. A tangent is drawn to the circle such that the segment of it inside the triangle is equal to $b$. Find the area of the triangle cut off by this tangent from the given one.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-245.jpg?height=462&width=1121&top_left_y=775&top_left_x=159)

Let $M N$ be the tangent to the circle inscribed in the given equilateral $\triangle A B C, M N=b$. Let

$$
\begin{gathered}
x=B M, y=B N . A M+C N=M N+A C=a+b \\
(\text { Theorem 2) } \Rightarrow x+y=(a-A M)+(a-C N)=a-b \Rightarrow \\
\Rightarrow(x+y)^{2}=(a-b)^{2}
\end{gathered}
$$

By the Law of Cosines for $\triangle B M N$,

$$
b^{2}=x^{2}+y^{2}-x y \Rightarrow
$$

$\Rightarrow(x+y)^{2}=b^{2}+3 x y=(a-b)^{2} \Rightarrow S_{B M N}=\frac{x y \sin 60^{\circ}}{2}=\frac{\left((a-b)^{2}-b^{2}\right) \sqrt{3}}{12}=\frac{a(a-2 b) \sqrt{3}}{12}$.

Answer: $\frac{a(a-2 b) \sqrt{3}}{12}$.