The vertices of square $A B C D$ are connected by a broken line $D E F G H B$. The smaller angles at vertices $E, F$, $G, H$ are right angles, and the segments $D E, E F, F G, G H, H B$ measure $6 \mathrm{~cm}, 4 \mathrm{~cm}, 4 \mathrm{~cm}, 1 \mathrm{~cm}$, $2 \mathrm{~cm}$ respectively.

Determine the area of square $A B C D$.

(M. Dillingerová)

![](https://cdn.mathpix.com/cropped/2024_04_17_e988acd5633232ee9496g-6.jpg?height=539&width=502&top_left_y=1995&top_left_x=777)

Hint. Could the actual diagonal be determined from that broken diagonal?
From the problem statement, we can determine the diagonal of the square, from which we can easily express its area.

The diagonal $D B$ is interpreted as the hypotenuse of the right triangle $D K B$, whose legs are parallel to the $s$ parts of the given broken line (the auxiliary quadrilaterals are rectangles):

![](https://cdn.mathpix.com/cropped/2024_04_17_e988acd5633232ee9496g-7.jpg?height=543&width=619&top_left_y=642&top_left_x=724)

The legs have the following lengths:

$$
\begin{aligned}
|D K| & =|D E|+|F G|+|H B|=6+4+2=12(\mathrm{~cm}) \\
|K B| & =|E F|+|G H|=4+1=5(\mathrm{~cm}) .
\end{aligned}
$$

According to the Pythagorean theorem, we have

$$
|D B|^{2}=|D K|^{2}+|K B|^{2}=12^{2}+5^{2}=169\left(\mathrm{~cm}^{2}\right) .
$$

The area of the square $A B C D$ is equal to half the square whose side is the diagonal of the square $A B C D$, i.e.,

$$
S_{A B C D}=\frac{1}{2}|D B|^{2}=84.5 \mathrm{~cm}^{2} .
$$

Note. The last relation can be considered well-known and does not need to be derived in the solution of the problem. It can be illustrated as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_e988acd5633232ee9496g-7.jpg?height=631&width=640&top_left_y=1900&top_left_x=708)

Alternatively (from the Pythagorean theorem in triangle $B A D$), we know that $|D B|^{2}=2|A B|^{2}$, thus

$$
|A B|^{2}=\frac{1}{2}|D B|^{2},
$$

which is precisely the area of the square $A B C D$.