77. (1985.VIII.3) We have (Fig. 49) $\overline{\mathrm{AP}}=3 \mathrm{~cm}$, $\overline{\mathrm{BP}}=8 \mathrm{~cm} \cdot \overline{\mathrm{OP}}=\overline{\mathrm{OM}}=\overline{\mathrm{ON}}=\mathrm{r}$

$\measuredangle C=90^{\circ}$. Therefore:

$\overline{\mathrm{AP}}=\overline{\mathrm{AN}}=3 \mathrm{~cm}-$

$\overline{B P}=\overline{B M}=8 \mathrm{~cm}$.

$\overline{C M}=\overline{C N}=\mathrm{r}$.

The area of the right triangle $\triangle \mathrm{ABC}$ is $P=\frac{1}{2} \overline{A C} \cdot \overline{B C}=\frac{1}{2} (r+3)(r+8)=$

$\frac{1}{2}\left(r^{2}+11 r+24\right)$

![](https://cdn.mathpix.com/cropped/2024_06_05_571619cb4100c1308092g-7.jpg?height=289&width=522&top_left_y=382&top_left_x=955)

Fig. 49

From the Pythagorean theorem for $\triangle \mathrm{ABC}$ we have:

$$
\begin{aligned}
& \overline{\mathrm{AC}}^{2}+\overline{\mathrm{BC}}^{2}=\overline{\mathrm{AB}}^{2} \\
& (r+3)^{2}+(r+8)^{2}=11^{2} \\
& r^{2}+6 r+9+r^{2}+16 r+64=121 \\
& 2 r^{2}+22 r=48 \\
& r^{2}+11 r=24
\end{aligned}
$$

If we substitute the result (2) into (1), we get:

$$
P=\frac{1}{2}(24+24)=24
$$

Answer: $P=24 \mathrm{~cm}^{2}$

Note. It can also be proven that the general statement is true: If $\mathrm{m}$ and n are the segments into which the hypotenuse is divided by the point of tangency of the inscribed circle in a right triangle, then its area is equal to $m \cdot n$.
Solution:

We have (see Fig. 49)

$P=\frac{1}{2}(m+r)(n+r)=\frac{1}{2}\left(m n+m r+n r+r^{2}\right)$.

From the Pythagorean theorem for $\triangle \mathrm{ABC}$ we have:

$$
\begin{aligned}
& (m+r)^{2}+(n+r)^{2}=(m+n)^{2} \\
& m^{2}+2 m r+r^{2}+n^{2}+2 n r+r^{2}=m^{2}+2 m n+n^{2} \\
& 2 r^{2}+2 m r+2 n r=2 m n \\
& \quad r^{2}+m r+n r=m n
\end{aligned}
$$

If we substitute the result (4) into (3), we get

$$
P=\frac{1}{2}(m n+m n)=m n
$$

which is what we needed to prove.