Area of equilateral triangle inscribed in right triangle

Rotate $\Delta Bps$ clockwise $60^\circ$ about $p$ as shown, to form $\Delta B'pq$.

Using $\Delta B'Cp$, find that $B'C=2\sqrt3~\mathrm{cm}$.

Using Sine Law on $\Delta B'Cq$, find that $\dfrac{B'q}{\sin 15^\circ}=\dfrac{B'C}{\sin 105^\circ}$, so $B'q=(4\sqrt3-6)~\mathrm{cm}$.

Use Pythagoras' theorem on $\Delta B'pq$ to find that $pq=2\sqrt{22-12\sqrt3}~\mathrm{cm}$.

Therefore the area is $\dfrac12a^2\sin60^\circ=(22\sqrt3-36)~\mathrm{cm}^2$.

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Geometry

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