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 $$DE = t$$. 

Tangents drawn to a circle from the same point are equal. Let $$T$$ be the point of contact of $$BE$$ with the semicircle

$$BA = BT, DE = DT$$

$$\implies BE = BA + DE = 2r + t$$

From the right angled triangle $$BCE$$ we get 

$$(2r + t)^2 = (2r)^2 + (2r - t)^2$$, 

which simplifies to $$t = \dfrac{r}{2}$$.

Then the area is $$\dfrac{1}{2} \times 2r \times \left(2r - \dfrac{r}{2}\right) = \dfrac{3r^2}{2} = 216$$

$$r^2 = 144 \implies r = 12$$.