3420.71 – Hard Triangle Area

If the side of the square is $a$ and $x$ is the height of the $\triangle RAC$, then the area of $\triangle RAC = \dfrac{ax}{2}$.

Now, $AD = x$ and $RD = x\sqrt3$, so $a = x + x\sqrt 3$ and $x = \dfrac{a}{\sqrt3 +1}$. This leads to

$$\begin{align*}\text{Area}(\triangle\text{RAC}) &= \dfrac{a}{2}\cdot\dfrac{a}{\sqrt3 +1} \\[1em]&= \dfrac{a^2}{2(\sqrt 3 +1)} \\[1em]&= \dfrac{a^2(\sqrt 3 - 1)}{2(\sqrt 3+1)(\sqrt 3-1)} \\[1em]&= \dfrac{a^2(\sqrt 3-1)}{2(3-1)} \\[1em]&= \dfrac{a^2(\sqrt 3-1)}{4}\end{align*}$$

And now, how about the area of $\triangle GCM$?