A circle is inscribed in a right triangle. The point of tangency divides the hypotenuse into two segments measuring 6 and $7 \mathrm{~cm}$. Calculate the area of this triangle.
Let $r$ be the radius of the circle inscribed in the triangle. Since the triangle is a right triangle, it is easy to verify that the legs of the triangle measure $6+r$ and $7+r \, \text{cm}$, as shown in the figure. By the Pythagorean Theorem, we have $(6+7)^{2} = (r+6)^{2}+(r+7)^{2}$, and expanding, we get $169=13^{2}=2\left(r^{2}+13 r\right)+85$, so that $r^{2}+13 r=42$.

Thus, the area of the triangle is

$$
\begin{aligned}
\frac{(r+6)(r+7)}{2} & =\frac{r^{2}+13 r+42}{2}=\frac{42+42}{2} \\
& =42 \, \text{cm}^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-074.jpg?height=306&width=439&top_left_y=1132&top_left_x=1274)