## Task B-1.4.

In trapezoid $A B C D$, $A B \| C D, |A B|=22, |B C|=10, |C D|=38, |D A|=14$. The angles of the trapezoid at vertices $A$ and $B$ are obtuse. The angle bisectors at vertices $A$ and $D$ intersect at point $E$, and the angle bisectors at vertices $B$ and $C$ intersect at point $F$. Calculate the area of the hexagon $A B F C D E$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d18d2bbdee068dba95dfg-03.jpg?height=388&width=1048&top_left_y=1391&top_left_x=493)

Let the angle $\varangle D A B$ be $2 \alpha$. Then the angle $\varangle D A E$ is $\alpha$. Due to the parallelism (bases of the trapezoid), we have $\varangle A D C = 180^{\circ} - 2 \alpha$, and thus $\varangle A D E = 90^{\circ} - \alpha$. Therefore, the angle $\varangle A E D = 90^{\circ}$. Similarly, we conclude that the angle $\varangle B F C = 90^{\circ}$.

Let $G$ be the intersection of $C D$ and $A E$ (the angle bisector at vertex $A$). Similarly, let $H$ be the intersection of $C D$ and $B F$.

The angles $\varangle D G A$ and $\varangle D A G$ are congruent (parallel legs), and the angles $\varangle G D E$ and $\varangle A D E$ are also congruent because $D E$ is the angle bisector. The side $\overline{D E}$ is a common side of triangles $G D E$ and $A D E$, so these triangles are congruent, making triangle $A G D$ an isosceles triangle. Similarly, we conclude that triangle $B C H$ is also isosceles. Therefore, the area of triangle $D E G$ is half the area of triangle $A D G$, and the area of triangle $B F C$ is half the area of triangle $B C H$.

The area of the hexagon $A B F C D E$ is:

$$
\begin{aligned}
& P_{A B F C D E} = P_{A B C D} - \frac{1}{2} P_{B C H} - \frac{1}{2} P_{A D G} \\
& = \frac{22 + 38}{2} \cdot h - \frac{1}{2} \cdot \frac{10 \cdot h}{2} - \frac{1}{2} \cdot \frac{14 \cdot h}{2} = \frac{1}{2} h (60 - 12) = 24 h
\end{aligned}
$$

where $h$ is the height of the given trapezoid $A B C D$.

We calculate the height from triangle $B C K$, which we obtain by drawing a parallel $B K$ to the leg $A D$ through vertex $B$. The lengths of the sides of triangle $B C K$ are $10, 16, 14$, so we calculate its area using Heron's formula:

$$
\begin{aligned}
P_{B C K} & = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2} \\
P_{B C K} & = 40 \sqrt{3}
\end{aligned}
$$

It follows that

$$
\begin{aligned}
& \frac{h \cdot 16}{2} = 40 \sqrt{3} \\
& h = 5 \sqrt{3}
\end{aligned}
$$

Thus, the desired area is $P_{A B F C D E} = 24 h = 24 \cdot 5 \sqrt{3} = 120 \sqrt{3}$.