# 2019 AMC 10B Problems/Problem 8

## Problem

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?

$(A)$ $4$ $(B)$ $12 - 4\sqrt{3}$ $(C)$ $3\sqrt{3}$ $(D)$ $4\sqrt{3}$ $(E)$ $16 - \sqrt{3}$

## Solution

We notice that the square can be split into 4 congruent smaller squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it HAS been split in 1/2).

When we split a equilateral triangle in 1/2, we get 2 triangles with a 30-60-90 relationship. Therefore, we get that the altitude and a side length of a square is sqrt 3 (Please help with LaTex!!).

We can then compute the area of the two triangles using the base-height-area relationship and get 2 * (sqrt 3) / 2 = sqrt 3.

The area of the small squares is the altitude square = sqrt 3 ^ 2 = 3. Therefore, the area of the shaded region in each of the four squares is (3 - sqrt 3).

Since there are four of these squares, we multiply this by 4 to get 4(3 - sqrt 3) = 12 - 4sqrt3 as our answer. This is choice $(B)$.

~ Awesome2.1

## See Also

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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