2015 AMC 8 Problem 2

Attempt Problem 2 of the 2015 AMC 8 below, then check your answer against the video solution and professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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}
Answer: D
Concepts:regular polygonarea decomposition
Small hint:
Draw segments from OO to every vertex of the octagon.
Big hint:
The shaded region contains three full central triangles and half of another.
Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

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 .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.