$\left[\begin{array}{l}\text { Pythagorean Theorem (direct and inverse) }\end{array}\right]$
[Area of a triangle (using two sides and the angle between them).]

Right triangles $A B C$ and $A B D$ have a common hypotenuse $A B=5$. Points $C$ and $D$ are located on opposite sides of the line passing through points $A$ and $B, B C=B D=3$. Point $E$ lies on $A C, E C=1$. Point $F$ lies on $A D, F D=2$. Find the area of the pentagon $E C B D F$.
## $S_{\triangle \mathrm{AEF}}=\frac{1}{2} A E \cdot A F \sin 2 \angle B A C$.

## Solution

From the right triangles $A B C$ and $A D B$, by the Pythagorean theorem, we find that $A C=A D=4$. Therefore, $A E=3$ and $A F=2$. Let $\angle B A C=\angle B A D=\alpha$. Then

$$
\sin \alpha=\frac{3}{5}, \cos \alpha=\frac{4}{5}
$$

Thus,

$$
S_{\triangle \mathrm{AEF}}=\frac{1}{2} A E \cdot A F \sin 2 \alpha=A E \cdot A F \sin \alpha \cos \alpha=3 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{72}{25}
$$

Therefore,

$$
S_{\mathrm{ECBDF}}=2 S_{\Delta \mathrm{ABC}}-S_{\Delta \mathrm{AEF}}=12-\frac{72}{25}=\frac{228}{25}=9.12
$$

## Answer

The line passing through points $G$ and $K$ serves as the angle bisector of $\angle F G H, K F \perp G F, K H \perp G H, K F=K H=8, G K=$

17. The segment $G L$ contains point $F$ and $F L=2$. The segment $G M$ contains point $H$ and $H M=19$. Find $L M$.

## Answer

$\sqrt{801}$.