<s>[INST] Find the ratio of the areas 
I'm having a trouble in solving this problem. How am I supposed to approach the problem?
 [/INST] Here's a "coordinate geometry" method:  Set up a Cartesian coordinate system so that A is (0, 0), B is (1, 0) and R is (0, -1).  Then Q is (1, -1).  Each interior angle of a regular hexagon is 120 degrees so angles FAR and BAS are both 120- 90= 30 degrees.  Then F is $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$  and S is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.   
From that, the two lines, through A and S and through F and P, have slope $-\frac{1}{\sqrt{3}}= -\frac{\sqrt{3}}{3}$.  In particular, the line through F and P can be written as $y+ \frac{\sqrt{3}}{2}= -\frac{1}{\sqrt{3}}\left(x- \frac{1}{2}\right)$$= -\frac{1}{\sqrt{3}}x+ \frac{1}{2\sqrt{3}}$ or $y= -\frac{1}{\sqrt{3}}x- \frac{1}{\sqrt{3}}$.
The lines through A and F and through S and P have slope $\sqrt{3}$.  In particular, the line through S and P can be written $y+ \frac{1}{2}= \sqrt{3}\left(x- \frac{\sqrt{3}}{2}\right)= x\sqrt{3}- \frac{3}{2}$.
P is the intersection of those two lines: $y= -\frac{1}{\sqrt{3}}x- \frac{1}{\sqrt{3}}= \sqrt{3}x- 2$ so $y= \frac{2\sqrt{3}- 1}{4}$, $x= \frac{2\sqrt{3}+ 5}{4\sqrt{3}}$.
Now that you know the coordinates of the vertices of each triangle, you can find the lengths of the sides and calculate the area using "Heron's formula": The area of a triangle with side lengths, a, b, and c, is given by $\sqrt{s(s- a)(s- b)(s- c)}$ where s is the "half perimeter, $s= \frac{a+ b+ c}{2}$.
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