Construct a circle with center $S$ and radius $3 \mathrm{~cm}$. Construct two mutually perpendicular diameters $A C$ and $B D$ of this circle. Construct isosceles triangles $A B K, B C L, C D M$, $D A N$ such that:

- the base of each triangle is a side of the quadrilateral $A B C D$,
- the base of each triangle is equal to the height to this base,
- no triangle overlaps the quadrilateral $A B C D$.

From the given data, calculate the area of the polygon $A K B L C M D N$. (M. Krejčová)
Analysis:

Segments $A C$ and $B D$ are congruent, perpendicular to each other, and the midpoint of each coincides with their intersection. Therefore, points $A, B, C, D$ form the vertices of a square. Triangle $A B K$ is isosceles with base $A B$, so its vertex $K$ lies on the perpendicular bisector of segment $A B$. The perpendicular bisector of segment $A B$ coincides with the perpendicular bisector of segment $C D$, and on this line lie the center $S$ of the square $A B C D$ and the vertex $M$ of the isosceles triangle $C D M$. Similar observations apply to the points $L, S$, and $N$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2298cc4d862d519ec02ag-2.jpg?height=737&width=826&top_left_y=1730&top_left_x=615)

Construction:

- Circle with center $S$ and radius $3 \mathrm{~cm}$,
- perpendicular lines passing through point $S$, intersections with the circle marked as $A, C$ and $B, D$,
- quadrilateral (square) $A B C D$,
- perpendicular to line $A B$ (identical to the perpendicular to line $C D$) passing through point $S$, foot of the perpendicular marked as $O$,
- point $K$ on the ray $S O$ at a distance $|A B|$ from point $O$,
- triangle $A B K$,
- other points and triangles similarly.

Calculation:

The polygon $A K B L C M D N$ is composed of the square $A B C D$ and four congruent isosceles triangles $A B K, B C L, C D M, D A N$.

Triangle $A B K$ has the base as a side of the square $A B C D$ and the corresponding height is equal to the side length. Therefore, the area of triangle $A B K$ is half the area of the square $A B C D$, and the area of the entire polygon $A K B L C M D N$ is three times the area of the square $A B C D$.

The square $A B C D$ is divided by its diagonals into four congruent triangles $A B S$, $B C S$, $C D S$, and $D A S$. Each of these triangles is right-angled and isosceles with legs of length $3 \mathrm{~cm}$. The area of the square is thus

$$
\frac{4 \cdot 3 \cdot 3}{2}=18\left(\mathrm{~cm}^{2}\right)
$$

The area of the polygon $A K B L C M D N$ is therefore $3 \cdot 18=54\left(\mathrm{~cm}^{2}\right)$.

Evaluation. 2 points for the analysis and construction; 4 points for the calculation, of which 2 points for determining the relationship between the areas of triangle $A B K$ and the square $A B C D$, 2 points for calculating the areas of the square and the entire polygon. A calculation based solely on measurement in the image, or a correct result without justification, is not evaluated.