13. On a sheet of paper, two regular hexagons are drawn. The smaller one has an area of 18, and a shorter diagonal of the larger hexagon coincides with a longer diagonal of the smaller hexagon. What is the area of the union of the two hexagons?
13. The answer is 29. The figure drawn on the paper can be represented as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_ffee1d4705848828ad5fg-16.jpg?height=566&width=691&top_left_y=134&top_left_x=728)

The point $O$ in the figure is the center of the larger hexagon $A B C D E F$. Since the hexagon $A B C D E F$ is regular, the triangle $O C D$ is equilateral. Furthermore, since $D A$ is the diameter of the circumscribed circle of $A B C D E F$, the triangle $A C D$ is a right triangle, with a right angle at $C$ and $\angle A D C = 60^{\circ}$. In particular, if $\ell$ is the length of the side $D C$ of the hexagon, then $A D = 2 \ell$. By the Pythagorean theorem, we have $A C = \sqrt{3} \ell$; it follows that the ratio between the major diagonal of the smaller hexagon $A K G C J I$ and the major diagonal of the hexagon $A B C D E F$ is $\frac{\sqrt{3}}{2}$, and therefore, the ratio of their respective areas is $\frac{3}{4}$, from which

$$
\operatorname{Area}(A B C D E F) = \frac{4}{3} \operatorname{Area}(A K G C J I) = \frac{4}{3} 18 = 24
$$

The trapezoid $A I J C$ is half of the hexagon $A K G C J I$, so

$$
\operatorname{Area}(A I J C) = \frac{1}{2} \operatorname{Area}(A K G C J I) = \frac{1}{2} 18 = 9.
$$

The quadrilateral $O C B A$ is one-third of the hexagon $A B C D E F$ (indeed, it is the union of the two equilateral triangles $O A B$ and $O B C$, both corresponding to $\frac{1}{6}$ of the hexagon); we obtain

$$
\operatorname{Area}(O C B A) = \frac{1}{3} \operatorname{Area}(A B C D E F) = \frac{1}{3} 24 = 8
$$

It is observed that the triangles $O A C$ and $B A C$ are congruent: indeed, $\angle C B A = 120^{\circ} = \angle C O A$, they have $C A$ as a common side, and both are isosceles; thus

$$
\operatorname{Area}(A B C) = \frac{1}{2} \operatorname{Area}(O C B A) = \frac{1}{2} 8 = 4.
$$

It follows directly that the area of the pentagon $A C D E F$ is given by

$$
\operatorname{Area}(A C D E F) = \operatorname{Area}(A B C D E F) - \operatorname{Area}(A B C) = 24 - 4 = 20.
$$

Finally, the area of the union of the two hexagons is given by the sum of the areas of the pentagon $A C D E F$ and the trapezoid $A I J C$, i.e.,

$$
\operatorname{Area}(A I J C D E F) = \operatorname{Area}(A I J C) + \operatorname{Area}(A C D E F) = 20 + 9 = 29.
$$