A convex quadrilateral $A B C D$ has an area of 20, and the coordinates of three of its vertices are $A(-2 ; 0), B(2 ; 0)$ and $C(2 ; 4)$. What is the minimum perimeter of the quadrilateral $A B C D$?
Solution. Based on the text, we draw a diagram. Since $t_{A B C}=\frac{4 \cdot 4}{2}=8$, it follows that $t_{A C D}=12$.

We know that

$$
t_{A C D}=\frac{A C \cdot m}{2}, \quad \text { and } \quad A C=4 \cdot \sqrt{2}
$$

thus $m=3 \cdot \sqrt{2}$.

Draw the line $e$ parallel to $A C$ and at a distance of $3 \cdot \sqrt{2}$ from $A C$. (There are two such lines, but only the one drawn in the problem is needed.) Reflect point $C$ over this line to get point $C^{\prime}$. The line $A C^{\prime}$ will intersect line $e$ at point $D$. We know that this is the shortest distance between $A$ and $C$ touching line $e$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1a211a5ab60e8c62d1c3g-1.jpg?height=577&width=585&top_left_y=552&top_left_x=759)

From the right triangle $A C C^{\prime}$:

$$
A C^{\prime 2}=(6 \cdot \sqrt{2})^{2}+A C^{2}=72+32=104, \quad A C^{\prime}=2 \cdot \sqrt{26}
$$

Then the perimeter of quadrilateral $A B C D$ is:

$$
\begin{aligned}
k & =A B+B C+C D+A D=A B+B C+A C^{\prime}=4+4+2 \cdot \sqrt{26}= \\
& =8+2 \cdot \sqrt{26} \approx 18.2
\end{aligned}
$$

Therefore, the minimum perimeter is approximately 18.2.