5. Given is the equilateral triangle $P Q R$. Inside this triangle, a regular hexagon $A B C D E F$ is drawn. Points $B, D$ and $F$ are the midpoints of the sides of triangle $P Q R$. The area of the pentagon $Q B A F R$ is equal to 1. What is the area of triangle $P Q R$?
A) $\frac{11}{10}$
B) $\frac{7}{6}$
C) $\frac{6}{5}$
D) $\frac{5}{4}$
E) $\frac{4}{3}$

![](https://cdn.mathpix.com/cropped/2024_04_17_4e6ce4fdc7c7bcbaf14eg-2.jpg?height=537&width=469&top_left_y=131&top_left_x=1339)
A5. C) $\frac{6}{5}$ We draw the center $M$ of the triangle, which is also the center of the hexagon. Draw line segments from $M$ to the vertices $B, F$, and $D$ of the hexagon. Also draw the line segments $A P, C Q$, and $E R$. Since the triangle is equilateral and the hexagon is regular, each of these six line segments is part of a line of symmetry of the triangle, which is also a line of symmetry of the hexagon. Finally, we draw the triangle $B F D$. $\mathrm{Now}$ we have divided the large triangle into twelve small triangles, as you can see in the figure below.

Due to the symmetry of the figure, all these small triangles are congruent: they are isosceles triangles with angles of $30^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_04_17_4472049c2281bf2d03f5g-2.jpg?height=534&width=465&top_left_y=1372&top_left_x=1341)
$30^{\circ}$, and $120^{\circ}$, and their base is always the same length as half the side of the large triangle. The gray pentagon has an area of 10 triangles, and the large triangle has an area of 12 triangles. Therefore, the area of the large triangle is $\frac{12}{10}=\frac{6}{5}$.