2019 AMC 10B Problem 8

The only thing we do not know yet is how big the square is. The triangles tell us.

1. Height of one triangle. Cut an equilateral triangle of side $2$ down the middle. You get a right triangle with hypotenuse $2$ and short leg $1$, so the height is $\sqrt{2^2 - 1^2} = \sqrt3$.
2. Size of the square. Each triangle sits on a side of the square and its top point is at the center. So the distance from the center to a side is $\sqrt3$. The side of the square is twice that: $2\sqrt3$.
3. Area of the square: $(2\sqrt3)^2 = 4 \cdot 3 = 12$.
4. Area of the triangles. One triangle has area $\frac12 \cdot 2 \cdot \sqrt3 = \sqrt3$. Four of them: $4\sqrt3$.
5. Shaded area: square minus triangles, $12 - 4\sqrt3$.

The answer is $\boxed{\textbf{(B)}\ 12 - 4\sqrt{3}}$.

In a figure made of several pieces, look for how one piece fixes the size of another before computing any area. Here the height of a triangle equals half the square's side because the tips meet at the center. After that, the shaded area is simply "big shape minus the pieces."

An equilateral triangle of side $2$ has height $\frac{\sqrt3}{2} \cdot 2 = \sqrt3$. Each triangle stands on a side of the square with its apex at the center, so the distance from the center to each side of the square is $\sqrt3$. The square therefore has side $2\sqrt3$ and area
$$(2\sqrt3)^2 = 12.$$

Each triangle has area $\frac{\sqrt3}{4} \cdot 2^2 = \sqrt3$, so the four triangles cover $4\sqrt3$. The shaded region is the square minus the triangles:
$$12 - 4\sqrt3.$$

The answer is $\boxed{\textbf{(B)}\ 12 - 4\sqrt{3}}$.