Ratio of area of 2 triangles in a hexagon 
I have no idea how to solve this question and it would be great if someone could help me with this.
 A: The answer is $\text{(D)}$.
Note that $\angle BAS=30^\circ$. 
This Implies that $\angle RAS=60^\circ$. Since $\overline {AR}=\overline {AS}$, $\triangle ARS$ is a regular triangle. 
Thus $\overline{RS}=\overline{AS}=\overline{SP}$. We conclude $\triangle {RSP}$ is a equilateral triangle where $\angle RSP=30^\circ$ and $SR=SP=1$. 
Also note that $\triangle {APQ}$ is a equilateral triangle with $AP=AQ=\sqrt{2}$, $\angle PAQ=30^\circ$. 
So we have that $\triangle {SRP} \sim \triangle {APQ} (\text{SAS})$. Since the ratio between $AP$ and $RS$ is $1:\sqrt{2}$, the ratio betwen the area of the two triangles will be $1:2$. 
A: Use symmetry to find the angles below, and the fact that squares have right angles while hexagons have $120^\circ$ as their interior angle.
Note that in triangle $SRP$, $SP=1$. Further, $RS=1$ because $\angle RAS=60^\circ$ and hence $AR=AS=RS$. Also, $\angle RSP=30^\circ$. Hence, area of $SRP = \frac{1}{2}*1 * 1* \sin 30^\circ = \frac{1}{4}.$
On the other hand, $AQ=\sqrt{2}$, and hence the area of triangle $APQ =\frac{1}{2} * \sqrt{2} *\sqrt{2} * \sin 30^\circ = \frac{1}{2}.$
Hence, the required ratio is  $\frac{1}{2}.$
