In the rhombus $A B C D$ with side length $4 \mathrm{~cm}$ and diagonal $A C$ of length $4 \mathrm{~cm}$, point $K$ is the midpoint of side $B C$. Furthermore, points $L$ and $M$ are constructed such that together with points $D$ and $K$ they form a square $K L M D$.

Calculate the area of the square $K L M D$.

(L. Růžičková)
According to the problem, the rhombus $A B C D$ is formed by two congruent equilateral triangles $A C D$ and $A B C$. Point $K$ is the midpoint of side $B C$, segment $A K$ is the height in triangle $A B C$ and the angle bisector of $\angle C A B$. The internal angles in the equilateral triangles have a size of $60^{\circ}$, so the size of $\angle C A K$ is $30^{\circ}$ and the size of $\angle D A K$ is $60^{\circ}+30^{\circ}=90^{\circ}$. Triangle $D A K$ is therefore a right triangle, with its hypotenuse being the side of the sought square.

Side $D A$ has a length of $4 \mathrm{~cm}$. Side $A K$ is a leg of the right triangle $A K C$, whose other sides measure $4 \mathrm{~cm}$ and $2 \mathrm{~cm}$. According to the Pythagorean theorem,

$$
|A K|^{2}=|A C|^{2}-|C K|^{2}=4^{2}-2^{2}=12\left(\mathrm{~cm}^{2}\right) .
$$

Now, according to the Pythagorean theorem in triangle $D A K$, we get

$$
|D K|^{2}=|D A|^{2}+|A K|^{2}=4^{2}+12=28\left(\mathrm{~cm}^{2}\right) .
$$

The area of square $K L M D$ is $28 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_9f120ead7c6455135b48g-3.jpg?height=911&width=759&top_left_y=1246&top_left_x=657)

Evaluation. 2 points for determining the right angle $D A K ; 2$ points for determining $|A K|^{2} ; 2$ points for determining $|D K|^{2}$.