Okay, I know I should have some work for my question, but I really have nothing.
We're supposed to find the area of a semicircle inscribed between the parabola 12-3x^2 and the x-axis. My teacher said we're supposed to use triangles to represent x and y values and the hypotenuse as our radius. I'm not really getting how to even begin solving it.

Well, I'm not sure what you have to find either. If you know how to sketch the graph of y = -3x^2 + 12, you'll have a general idea of how the semicircle is. If the semicircle has such a radius such that its centre is on the x axis, and it's circumference touches the curve, then you have a semicircle with centre (0, 0) with radius 2.

That seems too easy though...

y = -3x^2 + 12

So you get your maximum at (0, 12) and the x intercepts at -2 and 2.

No, it isn't a radius of 2. A radius of 2 actually crosses the parabola.

If we solve -3x^{2}+12=\sqrt{r^{2}-x^{2}} for x using the quadratic formula, we end up with a discriminant of 36r^{2}-143

Setting this to 0 and solving for r gives r=\frac{\sqrt{143}}{6}\approx 1.993

Just slightly less than 2.

I have attached the graph of the radius 2 case and this case to show the difference.

Oh, I get it now! Thanks galactus!

I had a feeling that was too easy the way I did ;)