2019 AMC 10B Problem 8

Question

The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length

2

and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

4

12 - 4\sqrt{3}

3\sqrt{3}

4\sqrt{3}

16 - 4\sqrt{3}

Answer: B

Concepts:area decomposition equilateral triangle special right triangle

Difficulty rating: 1330

Small hint:

Find the altitude of each equilateral triangle

Big hint:

Subtract the four triangle areas from the square’s area

Solution:

Each equilateral triangle has side length

2

, so its altitude is

\sqrt3

. Because that altitude runs from a side of the square to its center, the square has side length

2\sqrt3

and area

12

.

Each of the four equilateral triangles has area

\frac{\sqrt3}{4}(2^2)=\sqrt3

. Therefore the shaded area is

12-4\sqrt3.

Thus, the answer is B .