Area of the overlap between a triangle and a square $ABC$ is an equilateral triangle, each side has length 4.  
$M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.
I'm not sure how to solve this problem, or even where to start. Any hints?
 A: Use this diagram.

The equilateral triangle $ABC$ has sides of length $4$, so $CM=2$ and $AM=2\sqrt 3$.
The area you are trying to find is shaded and is clearly twice the area of triangle $AEM$. We know the base, $AM$, so we want the height $EF$. If we let $x=EF$ then $MF=x$ due to $45-45-90$ triangle $EFM$. We also have $AF=\sqrt 3x$ due to $30-60-90$ triangle $AEF$. Therefore we get
$$AF+MF=AM$$
$$\sqrt 3x+x=2\sqrt 3$$
$$x=\frac{2\sqrt 3}{\sqrt 3+1}$$
$$=\frac{2\sqrt 3(\sqrt 3-1)}{(\sqrt 3+1)(\sqrt 3-1)}$$
$$=\frac{6-2\sqrt 3}{2}$$
$$=3-\sqrt 3$$
Therefore the area we want is twice the area of triangle $AEM$:
$$Area=2\cdot \frac 12bh$$
$$=2\cdot \frac 12 \cdot 2\sqrt 3 \cdot (3-\sqrt 3)$$
$$=6\sqrt 3-6$$
A: I'm assuming that this is the figure you mean.
                         
First off, notice that the overlapping area is a kite shape, and since kites are orthodiagonal quadrilaterals, we can find the area using the formula
$$
A=\frac{p\cdot q}{2}
$$
Where $p$ and $q$ are the kite's diagonals. The long diagonal is clearly just the height of the triangle, which is just
$$r\cos\:(30^\circ)=\frac{4\sqrt {3}}{2}=p$$
The shorter diagonal is a bit more challenging to find. The easiest to understand, in my opinion, is to find the positions where the sides of the square intersect the sides of the triangle (you only need to find one). This can be solved as a system of equations, since we know the angles from the center diagonal of the triangle and square to be $30^\circ$ and $45^\circ$ respectively:
$$
a\sin (45^\circ) = b\sin (30^\circ)
$$$$
a\cos (45^\circ) = -b\cos (30^\circ)
$$
Once you've solved for $a$ and $b$, the second diagonal of the kite is just
$$
2a\sin (45^\circ)=q
$$ or $$
2b\sin (30^\circ)=q
$$
Hope this helped :-)
