The polygons are regular polygons. Find the area of the shaded region.

Answer:[tex]194.8557 in^2[/tex]The side of a regular hexagon is congruent with the radius of the circle the hexagon is inscribed with.  That makes the two triangles in red and green (lines are a bit offset for clarity) equilateral. At this point it's easy to compute the area of an equilateral triangle known the side length l, which is [tex]A = \frac12 l (l\frac{\sqrt3}2) = \frac{\sqrt3}4 l^2[/tex] At this point, the area of the shaded region of one triangle making up the hexagon is obtained by difference (area of the red triangle, minus area of the white:[tex]A=A_r-A_w = \frac{\sqrt3}4 10^2 -\frac{\sqrt3}45^2 = \frac{\sqrt3}4(10^2-5^2)=\frac{\sqrt3}4\times 75\approx 32.47595[/tex]The area of the whole shaded region, is 6 times that, or [tex]194.8557[/tex]With tasselation:Join the midpoint of the side of the external hexagon with the end points of the corresponding side of the inner hexagon. You have now created four equilateral squares which are congruent. (yes, my drawing skills are terrible)The shaded region is made of 18 triangular tiles, each having an area of (see above formula) [tex]\frac{\sqrt3}4\times 5^2 = 18.8253[/tex], for a total of, again, 194.8557 square inch.