13) The ratio between the area of the regular hexagon and that of the star polygon shown in the figure, which has all its sides lying on 6 of the diagonals of the hexagon, is
(A) $\frac{4}{3}$
(B) $\frac{3}{2}$
(C) $\frac{5}{3}$
(D) $\frac{6}{5}$
(E) $\frac{5}{4}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ad8c4c2d860e923409d8g-2.jpg?height=232&width=257&top_left_y=354&top_left_x=1147)
13) The answer is (B)

By drawing the lines indicated in the figure, it is easily verified that the hexagon is divided into 18 equivalent triangles. Indeed, the diagonals emanating from each vertex divide the diagonal connecting the two adjacent vertices to the chosen one into 3 equal parts, and the six obtuse triangles have the same base and the same height as the equilateral triangles. Since the area of the star polygon is given by the sum of the areas of the 12 equilateral triangles (all equal to each other), the ratio of the areas is $\frac{18}{12}=\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_855938a0fb7ccfca2bc5g-3.jpg?height=220&width=260&top_left_y=353&top_left_x=955)