1. Given a right triangle $A B C$ with a right angle at vertex $C$. If the length of one leg is increased by $7 \mathrm{~cm}$, while the other leg remains the same, the hypotenuse of the new triangle would be $5 \mathrm{~cm}$ longer than the original hypotenuse. If the length of the same leg in the original triangle is decreased by $4 \mathrm{~cm}$, while the other leg remains the same, the new hypotenuse would be $2 \mathrm{~cm}$ shorter than the original hypotenuse. Calculate the area of the original triangle $A B C$.
1. 

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-16.jpg?height=862&width=1062&top_left_y=731&top_left_x=494)

Let $a, b$, and $c$ be the standard notations for the lengths of the sides of the right triangle $ABC$.

The Pythagorean theorem states:

$$
a^{2}+b^{2}=c^{2}
$$

Applying the Pythagorean theorem to the right triangle $\mathrm{A}_{1} B C$, with side lengths $a, b+7$, and $c+5$, we get:

$$
a^{2}+(b+7)^{2}=(c+5)^{2}
$$

Applying the Pythagorean theorem to the right triangle $A_{2} B C$, with side lengths $a, b-4$, and $c-2$, we get:

$$
a^{2}+(b-4)^{2}=(c-2)^{2}
$$

After squaring the binomials, we obtain the system of equations:

$$
\begin{aligned}
& a^{2}+b^{2}=c^{2} \\
& a^{2}+b^{2}+14 b+49=c^{2}+10 c+25 \\
& a^{2}+b^{2}-8 b+16=c^{2}-4 c+4
\end{aligned}
$$

From the first two equations, by subtracting (1*) from $\left(2^{*}\right)$, or by substituting $c^{2}=a^{2}+b^{2}$, we get

$$
\begin{aligned}
14 b-10 c & =-24 /: 2 \\
7 b-5 c & =-12
\end{aligned}
$$

From the first and third equations, similarly, we get

$$
\begin{gathered}
-8 b+4 c=-12 /: 4 \\
-2 b+c=-3
\end{gathered}
$$

Now, from the system of equations (4) and (5), $b$ and $c$ can be calculated:

$$
\begin{gathered}
7 b-5 c=-12 \\
-2 b+c=-3 / \cdot 5 \\
\hline 7 b-5 c=-12 \\
-10 b+5 c=-15 \\
-3 b=-27 \Rightarrow b=9 \mathrm{~cm} .
\end{gathered}
$$

From (5) it follows: $c=2 b-3=18-3=15 \mathrm{~cm}$.

Then, $a^{2}=15^{2}-9^{2}=225-81=144 \Rightarrow a=12 \mathrm{~cm}$.

The area of triangle $ABC$ is: $P=\frac{a \cdot b}{2}=\frac{12 \cdot 9}{2}=54 \mathrm{~cm}^{2}$.