Find the area of the trapezoidal surface $ABCD$

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$

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