2019 AMC 10B Problem 8

Attempt Problem 8 of the 2019 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10B solutions, or check the answer key.

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8.

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 22 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 4

1243 12 - 4\sqrt{3}

33 3\sqrt{3}

43 4\sqrt{3}

1643 16 - 4\sqrt{3}

Answer: B
Concepts:area decompositionequilateral trianglespecial right triangle
Difficulty rating: 1330
Solution:

Each equilateral triangle has side length 22, so its altitude is 3\sqrt3. Because that altitude runs from a side of the square to its center, the square has side length 232\sqrt3 and area 1212.

Each of the four equilateral triangles has area 34(22)=3\frac{\sqrt3}{4}(2^2)=\sqrt3. Therefore the shaded area is 1243.12-4\sqrt3.

Thus, the answer is B .

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