2019 AMC 10B Problems/Problem 8

Revision as of 02:09, 16 February 2019 by Awesome2.1 (talk | contribs) (Solution)


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?

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


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 already been split in half).

When we split an equilateral triangle in half, we get $2$ triangles with a $30-60-90$ relationship. Therefore, we get that the altitude and a side length of a square is $\sqrt{3}$.

We can then compute the area of the two triangles using the base-height-area relationship and get $2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}$.

The area of the small squares is the altitude squared which is $(\sqrt{3})^2 = 3$. 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 $4(3 - \sqrt{3}) = 12 - 4 \sqrt{3}$ as our answer. This is choice $\boxed{ B) 12 - 4 \sqrt{3}}$.

~Awesome2.1 and edited by greersc.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 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

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