1992 AMC 8 Problem 24

Attempt Problem 24 of the 1992 AMC 8 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 1992 AMC 8 solutions, or check the answer key.

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

24.

Four circles of radius 33 are arranged as shown. Their centers are the vertices of a square. The area of the shaded region is closest to
7.77.7
12.112.1
17.217.2
1818
2727
Answer: A
Concepts:circle areaarea decompositionestimation
Small hint:
Since neighboring circles are tangent, the square’s side equals two radii, so its side is 66 and its area is 3636
Big hint:
The four quarter-circles inside the square together make one full circle of area 9π9\pi; subtract that from the square
Solution:
Because adjacent circles are tangent, the square through the centers has side 2×3=62 \times 3 = 6 and area 36.36.
Inside the square, each circle contributes a quarter-circle, and the four quarters make one full circle of area 9π≈28.3.9\pi \approx 28.3. The shaded region is 36−9π≈7.7.36 - 9\pi \approx 7.7.
Thus, the correct answer is A .
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.