2015 AMC 8 Problem 2

Below is the video solution and professionally curated solution for Problem 2 of the 2015 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 8 solutions, or check the answer key.

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Concepts:regular polygonarea decomposition

Difficulty rating: 1020

2.

Point OO is the center of the regular octagon ABCDEFGH,ABCDEFGH, and XX is the midpoint of the side AB.\overline{AB}. What fraction of the area of the octagon is shaded?

1132\dfrac{11}{32}

38\dfrac{3}{8}

1332\dfrac{13}{32}

716\dfrac{7}{16}

1532\dfrac{15}{32}

Video solution:
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Written solution:

First notice that there are 88 equally sized triangles that can be created with OO and any two consecutive points. Therefore, they each take up 18\dfrac {1}{8} of the total area of the octagon.

The shaded area has three complete triangles and half of the triangle ABO.ABO. Therefore, the shaded area is 3.58=716\dfrac{3.5}{8} = \dfrac{7}{16} of the total area of the octagon.

Thus, the correct answer is D .

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