1987 AMC 8 Problem 22

Attempt Problem 22 of the 1987 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 1987 AMC 8 solutions, or check the answer key.

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

22.

ABCDABCD is a rectangle, DD is the center of the circle, and BB is on the circle. If AD=4AD = 4 and CD=3,CD = 3, then the area of the shaded region is between which two values?
44 and 55
55 and 66
66 and 77
77 and 88
88 and 99
Answer: D
Concepts:Pythagorean Theoremsector
Small hint:
The diagonal DBDB is the radius; find it with the Pythagorean theorem
Big hint:
The shaded region is a quarter of the circle with the rectangle removed: 14π(5)2−3⋅4\dfrac14 \pi (5)^2 - 3 \cdot 4
Solution:
Since AD=4AD = 4 and CD=3,CD = 3, the diagonal DB=32+42=5,DB = \sqrt{3^2 + 4^2} = 5, which is the radius.
The shaded region is the quarter circle at DD minus the rectangle: 14π(5)2−(3)(4)=25π4−12≈19.6−12=7.6. \begin{gathered} \dfrac14 \pi (5)^2 - (3)(4) = \dfrac{25\pi}{4} - 12 \\ \approx 19.6 - 12 = 7.6. \end{gathered}
This lies between 77 and 8.8.
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.