Calculate the area of 2 triangles within the square $ABCD$ is a square with a side of length 4. P is on AB, S is on CD and Q is on PS such that:


*

*$AP = CS$

*The triangles $PBR$ and $SDQ$ are both equilateral triangles. See the image below. 

Calculate the combined area of the 2 triangles.
What would be the easiest way the solve this? I first tried to name SC x, and then make an equality involving x (something equalling 16, the area of the square, and then solving for that x) but that didn't work out. 
 A: Let's do it in the style that you did not complete. While it is not the easiest way, it is completely mechanical. 
We write down faithfully the first calculation that we did. Then we describe a much more streamlined way of doing the same thing. 
Let $a=AP=CS$. We find the equation of the "left" side of the bottom triangle. This passes through $(a,0)$ and  has slope $\tan(60^\circ)=\sqrt{3}$.
So the equation is 
$$\frac{y-0}{x-a}=\sqrt{3},$$
or equivalently $y=\sqrt{3}x-a\sqrt{3}$.
This line is the same as the "right" side of the upper triangle.
We find the equation of that right side. It has slope $\sqrt{3}$, and passes through the point $(4,4-a)$. So the equation of the right side of the upper triangle is
$$\frac{y-4}{x-4+a}=\sqrt{3},$$
or equivalently $y=\sqrt{3}x +4-4\sqrt{3}+a\sqrt{3}$.
Compare the two equations we have obtained. Their constant terms must be the same. This gives us a linear equation for $a$.
Another way:  We want the line through $(a,0)$ and $(4-a,4)$ to have slope $\sqrt{3}$. That yields the equation 
$$\frac{4}{4-2a}=\sqrt{3}.$$
It is slightly more convenient to let the side of the triangles be $b$. then we are talking about the line from $(4-b,0)$ to $(b,4)$. We get the equation
$$\frac{4}{2b-4}=\sqrt{3}.$$
Now that we have $b$, we can find the areas of the triangles by using a standard formula. 
