Question

Find the area of the shaded region. Round to the nearest tenth. All polygons are regular.

Answer 1

Check the picture below.

let's take a looksie at that, notice, if we split the circle into 6 even pieces, we end up with six 60° angles at the center, namely six equilateral triangles as you see there.

now, if we run a dashed line as you see from a corner to the other to make the flatter triangle which is shaded, we end up cutting the equilateral triangle into two equal halves, each with 30° and 30° angles, because the dashed line intercepts the green line at 90°.

well, the flatter triangle from corner to corner, is spanning over two equilateral triangles, if it's owning half of one, it must be also owning half of the other. What the dickens does that mean? well, it means the area of the flatter triangle that's shaded, is really the same area as a "whole" of one of those equilateral triangles.

so hmmm we have three of those flatter triangles shaded hmm, so let's simply get the area of three of those equilateral triangles and sum them up.

`\textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2√(3)}{4} ~~ \begin{cases} s=side\\[-0.5em] \hrulefill\\ s=12 \end{cases}\implies A=\cfrac{12^2√(3)}{4}\implies A=36√(3) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large area of the shaded region}}{36√(3)~~ + ~~36√(3)~~ + ~~36√(3)}\implies 108√(3)\qquad \approx\qquad \text{\LARGE 187.1}`