If the area of the region bounded by the curves $$y = 4 - \frac{x^2}{4}$$ and $$y = \frac{x-4}{2}$$ is equal to $$\alpha$$, then $$6\alpha$$ equals

We need the area bounded by $$y = 4 - \frac{x^2}{4}$$ and $$y = \frac{x-4}{2}$$.

Finding intersection points by setting the curves equal:

$$4 - \frac{x^2}{4} = \frac{x-4}{2}$$

Multiplying by 4: $$16 - x^2 = 2(x-4) = 2x - 8$$

$$x^2 + 2x - 24 = 0$$, giving $$(x+6)(x-4) = 0$$, so $$x = -6$$ or $$x = 4$$.

The area is:

$$\alpha = \int_{-6}^{4} \left[\left(4 - \frac{x^2}{4}\right) - \frac{x-4}{2}\right] dx = \int_{-6}^{4} \left(6 - \frac{x}{2} - \frac{x^2}{4}\right) dx$$

$$= \left[6x - \frac{x^2}{4} - \frac{x^3}{12}\right]_{-6}^{4}$$

At $$x = 4$$: $$24 - 4 - \frac{64}{12} = 20 - \frac{16}{3} = \frac{44}{3}$$

At $$x = -6$$: $$-36 - 9 + 18 = -27$$

$$\alpha = \frac{44}{3} - (-27) = \frac{44}{3} + \frac{81}{3} = \frac{125}{3}$$

$$6\alpha = 6 \times \frac{125}{3} = 250$$

Hence, the correct answer is Option A.