In a regular hexagon with an area of $144 \mathrm{~cm}^{2}$, we have marked all its diagonals. The hexagon thus "broke down" into several triangles and quadrilaterals.

a) Determine into how many parts the diagonals divided the hexagon.

b) Determine the area of the regular hexagon that will be formed by combining all the quadrilateral parts of the given hexagon.

(Bednářová)
First, let's draw the entire situation.

![](https://cdn.mathpix.com/cropped/2024_04_17_3fba1832e0e68708cfa9g-6.jpg?height=374&width=377&top_left_y=838&top_left_x=885)

Fig. 1

From the first image, it is clear that the hexagon is composed of six triangles similar to triangle \( SAB \).

a) Given that triangle \( SAB \) consists of 4 parts, we see that the diagonals divide the hexagon into 24 parts (3 points).
b)

![](https://cdn.mathpix.com/cropped/2024_04_17_3fba1832e0e68708cfa9g-6.jpg?height=404&width=414&top_left_y=1554&top_left_x=861)

Fig. 2

From Fig. 2, it follows that triangle \( SAB \) can be divided into 6 congruent smaller triangles. Since two of these are part of the hexagon whose area we need to determine, we see that the area of the smaller hexagon is one-third of the area of the original hexagon. Therefore, the area of the hexagon we are looking for is \( 144 \div 3 = 48 \, \text{cm}^2 \).

(Full solution: 6 points)