An equitaleral triangle is inscribed in a circle...

swerve wrote:

An equilateral triangle is inscribed in a circle, as shown above. What is the area of the shaded region, if the area of the circle is 2?

$$A.\ \ 2-3\sqrt{3}$$
$$B.\ \ 2-3\sqrt{3}\pi$$
$$C.\ \ 2-\frac{3\sqrt{3}}{4}$$
$$D.\ \ 2-\frac{3\sqrt{3}}{2\pi}$$
$$E.\ \ 2\pi-\frac{3\sqrt{3}}{2\pi}$$

The OA is D.

Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.

Well, we know the answer is going to be the area of the circle - area of the equilateral triangle, so it has to be 2 - something. Eliminate E.

Notice that A and B would both yield negative answers, as we're calculating 2 - some value greater than 2, so those are out.

Last, if the area of the circle is 2, then we can find the radius. But all we really need to see is that there's going to be a Pi term in the radius. If that's not clear, you can quickly solve for the radius.
2 = Pi * r^2
r^2 = 2/Pi
r = rt(2/Pi)

If we're going to use the radius to find a side of the triangle, and there's a Pi term in the radius, then there's going to have to be a Pi term in the triangle as well. There's no Pi term in C.

We're left with D. No need to do any math at all.