\section*{Task 4 - 181014}

Let there be a triangle \(A B C\) with a right angle \(A C B\).

The inradius is \(\rho\).

(It is often called this way.)

Then let \(c\) also be known.

(It can also be called the hypotenuse length.)

Now it is necessary to express the area using only these pieces.

One must adhere to the laws, (but one does not need to rhyme while doing so.)
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1404.jpg?height=414&width=559&top_left_y=824&top_left_x=748)

Let the incenter be \(O\).

Then the radius \(\rho\) is certainly perpendicular to \(A B\) once, and secondly it is perpendicular to \(B C\).

Also with \(A C\), the third segment, it forms a right angle;

The points where it intersects are denoted by \(D, E, F\).

The triangle \(A B O\) has an area of \(c \cdot \rho: 2\)

which is further divided by \(D O\). When we consider \(\triangle A D O\) and compare it with \(\triangle A F O\), we see - as is often "easy" - that both areas are congruent if we know the congruence theorem described by ssw, where \(A O\) is equal to itself; then \(C D\) is exactly equal to \(\rho\) and the same applies to \(O F\); the angle \(\angle A D O\) is a right angle, and \(\angle A F O\) is no less.

The same applies to the triangle \(B O D\) when compared with \(\triangle B O E\). The quadrilateral FCEO remains as the rest. Now we think as follows:

Since three of the angles are 90 degrees, the fourth must also be 90 degrees. Therefore, it must be a rectangle. When we substitute the side lengths, we recognize that it consequently has an area of \(\rho^{2}\).

And our solution is now as follows:

\[
\rho^{2}+c \cdot \rho
\]

\subsection*{7.20.2 II. Round 1978, Class 10}