For reference:
In the figure shown, calculate the area of the trapezoidal surface $ABCD$, if $\overset{\LARGE{\frown}}{CD} = 150^o$and $BM=MA$. (Answer:$R^2$)
Let $R = 1, AM=BM = x:$
$\sin(15°)=\frac{MC}{2}\implies MC=\sqrt{2}-\sqrt{3}$
$\tan(15°)=\frac{BC}{2x}\implies BC=2x(2-\sqrt3)$
$\triangle BMC: x^2=\frac{10+3\sqrt 3}{73}$
$S=\frac{(BC+AD)×AB}{2}=x^2(4-2\sqrt3)+x\sqrt{4-x^2}$
$\implies=\frac{22-8\sqrt 3}{73}+\frac{51+8\sqrt 3}{73}=1=R^2$
Is there another way of solving? Without considering $R = 1$?
Here is my solution that avoids much calculation -
$ \small [BCDM] = [BCD] + [BDM] = [BCA] + [ADM]$
$ \small = 2 [BCM] + [ADM]$
But also, $ \small [BCDM] = [BCM] + [CDM] \implies [ADM] + [BCM] = [CDM]$
$ \small \implies [BCDA] = [ADM] + [BCM] + [CDM] = 2 [CDM]$
Drop a perp from $ \small C$ to $ \small DM$, which is $ \small R/2$.
$ \small \displaystyle [CDM] = \frac 12 \cdot \frac R 2 \cdot 2R = \frac{R^2}{2}$
$ \therefore \small [BCDA] = R^2$