ABCD is a square. BE is the tangent to the semicircle on AD as diameter. The area of the triangle BCE is $$216 \text{ cm}^2$$. The radius of the semicircle (in cm ) is

Let the side of the square be $$2r$$ and $$DE = t$$. Equal tangents give $$BE = BA + DE = 2r + t$$, and the right triangle $$BCE$$ gives $$(2r + t)^2 = (2r)^2 + (2r - t)^2$$, which simplifies to $$t = \frac{r}{2}$$. Then the area is $$\frac{1}{2} \times 2r \times \left(2r - \frac{r}{2}\right) = \frac{3r^2}{2} = 216$$, so $$r^2 = 144$$ and $$r = 12$$.