A circle is inscribed in a right triangle. The point of tangency $\mathrm{O}$ divides the hypotenuse into two segments of lengths $6 \mathrm{~cm}$ and $7 \mathrm{~cm}$. Calculate the area of the triangle.
Let $r$ be the radius of the inscribed circle. Using the Pythagorean theorem, we have $(6+7)^{2}=(r+6)^{2}+(r+7)^{2}=r^{2}+12 r+36+r^{2}+14 r+49=2\left(r^{2}+13 r\right)+85$, thus we have $r^{2}+13 r=\frac{169-85}{2}=42$.

On the other hand, the area of the triangle is

$\frac{(r+6)(r+7)}{2}=\frac{r^{2}+13 r+42}{2}=\frac{42+42}{2}=42$.

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