In the figure below, $A B C D$ and $E F G H$ are rectangles with parallel sides. Given that $A E=10$, $B F=20$ and $D H=30$, determine the length of the segment $C G$.

![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-29.jpg?height=471&width=751&top_left_y=581&top_left_x=755)
Solution

We will perform two geometric transformations on the drawing to keep the lengths of $A E, D H, G C$ and $B F$ unchanged. Translate ${ }^{1}$ the trapezoid $A E F B$ to the right as indicated in the figure below until $E F$ coincides with $H G$. Then, translate the triangle $\triangle A H D$, also as indicated below, until $H$ coincides with $G$.
![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-29.jpg?height=952&width=1354&top_left_y=1552&top_left_x=455)

Let $P$ be the new point obtained by the collapse of $E, H, G$ and $F$, and let $x, y, z$ and $w$ be their distances to the sides of the rectangle. By the Pythagorean Theorem, we have:

$$
\begin{aligned}
& A E^{2}=A P^{2}=x^{2}+z^{2} \\
& D H^{2}=P D^{2}=x^{2}+w^{2} \\
& G C^{2}=P C^{2}=y^{2}+w^{2} \\
& B F^{2}=P B^{2}=z^{2}+y^{2}
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
A E^{2}+G C^{2} & =x^{2}+y^{2}+z^{2}+w^{2} \\
& =D H^{2}+B F^{2} \\
& =900+400
\end{aligned}
$$

Finally, $G C=\sqrt{900+400-100}=20 \sqrt{3}$.