The area of the right triangle ABC ( $\angle C=90^{\circ}$ ) is 6, and the radius of the circumscribed circle around it is $\frac{5}{2}$. Find the radius of the circle inscribed in this triangle.

#
The radius of the circle inscribed in a right-angled triangle with legs $a, b$ and hypotenuse $c$ is $\frac{a+b-c}{2}$.

## Solution

It is known that the radius $r$ of the circle inscribed in a right-angled triangle with legs $a, b$ and hypotenuse $c$ is $\frac{a+b-c}{2}$, and the area is $\frac{ab}{2}$.

Since the center of the circle circumscribed around a right-angled triangle coincides with the midpoint of the hypotenuse, then $c=5$. By the Pythagorean theorem, $a^{2}+b^{2}=25$. Therefore,

$$
(a+b)^{2}=a^{2}+b^{2}+2ab=25+4 \cdot 6=49, a+b=7
$$

Thus,

$$
r=\frac{a+b-c}{2}=\frac{7-5}{2}=1
$$

## Answer

1.