Find the area of the region between a regular hexagon with sides of 6" and its inscribed circle.

You can compute the area of the region between the hexagon and the circle by subtracting:

Area of region = Area of hexagon - Area of inscribed circle

Let's find those areas.

If we draw a segment from the center of the hexagon to each of its vertex, the hexagon will be divided into 6 equal triangles. For any of those triangles, the angle in the center equals 360°/6 = 60°. Being the other angles equal, all three angles must equal 60° (because they must add up to 180°). Hence, those triangles are equilateral. The height (h) from the center divides the triangle into two equal right triangles. We know the hypotenuse is 6, because of the equilateral triangle we began with, and one of the other sides is half the side of the hexagon, 3. Applying the Pythagorean Theorem to this right triangle, we find that its height is:
 
[tex]h = \sqrt{6^2-3^2}=\sqrt{27} = 3\sqrt{3} = 5.20[/tex]

By the way, this height is also the radius of the inscribed circle, and we'll be using it later to find its area.

Now, we can find the area of each of the six triangles into which we divided the hexagon:

[tex]A = \dfrac{1}{2}\cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3} = 15.59[/tex]

There are six of those triangles. So, finally, the area of the hexagon is:

Area of hexagon = 6A = [tex]54\sqrt{3} = 93.53[/tex]

Let's move on to the circle. It's easier, because we already know its radius. The area of a circle with radius r is given by:

Area of circle = [tex]\pi\cdot r^2 = \pi \cdot \left(3\sqrt{3}\right)^2 = 27\cdot \pi = 84.82[/tex]

Last, the area of the region:

Area of region = Area of hexagon - Area of circle = 93.53 - 84.82 = 8.71