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? \t\t

$4$

$12 - 4\sqrt{3}$

$3\sqrt{3}$

$4\sqrt{3}$

$16 - 4\sqrt{3}$

Solution:

The altitude of the triangle is $\sqrt 3$ using $30-60-90$ triangles, so the total base is $2\sqrt 3.$ The total amount of the base on each side that isn't in the white region is $2\sqrt 3 -2,$ so the amount from each triangle is $\sqrt 3 -1.$

This makes $8$ total triangles with base $\sqrt 3-1$ and altitude $\sqrt 3,$ so the combined area is $8\cdot \dfrac{(\sqrt 3)(\sqrt 3-1)}2$ $= 4\cdot (3-\sqrt 3)$ $= 12-4\sqrt 3.$

Thus, the answer is B .

2019 AMC 10B Problem 8

Solution:

Problem 8 in Other Years