Let $O$ be the center of the circle (the midpoint of $A B$). Represent the area of the quadrilateral $A E G F$ as the sum of the areas of triangles $A O F, G O E, A O E$, and $G O F$.

## Solution

The center $O$ of the circle circumscribed around the right triangle $A B C$ coincides with the midpoint of the hypotenuse $A B$. Therefore, the radius of this circle is half the hypotenuse, i.e., $R=\frac{13}{2}$.

The perpendicular bisectors of the legs $A C$ and $B C$ intersect the smaller arcs $A C$ and $B C$ at their midpoints $E$ and $G$, respectively. Therefore,

$$
\angle E O A = \angle C B A, \angle G O E = 90^{\circ}.
$$

The perpendicular bisector of the hypotenuse $A B$ intersects the arc $A B$, not containing point $C$, also at its midpoint $F$. Therefore,

$$
\angle A O F = 90^{\circ}, \angle G O F = 180^{\circ} - \angle C B A.
$$

Consequently,

$$
\begin{gathered}
S_{\mathrm{AEGF}} = S_{\triangle \mathrm{AOF}} + S_{\Delta \mathrm{GOE}} + S_{\triangle \mathrm{AOE}} + S_{\Delta \mathrm{GOF}} = \\
= \frac{1}{2} R^{2} + \frac{1}{2} R^{2} + \frac{1}{2} R^{2} \sin \angle B + \frac{1}{2} R^{2} \sin \left(180^{\circ} - \angle B\right) = R^{2} + \frac{5}{13} R^{2} = \frac{117}{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-16.jpg?height=488&width=509&top_left_y=2157&top_left_x=777)

## Answer

$\frac{117}{2}$.
A circle is circumscribed around a right triangle $A B C$ with legs $A C=5$ and $B C=12$. Points $E$ and $G$ are the midpoints of the smaller arcs $A C$ and $B C$ of this circle, and point $F$ is the midpoint of the arc $A B$ that does not contain point $C$.

Find the area of the quadrilateral $A E G F$.

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