2004 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2004 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

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Concepts:circle areasectorinclusion-exclusion

Difficulty rating: 1270

9.

A square has sides of length 10,10, and a circle centered at one of its vertices has radius 10.10. What is the area of the union of the regions enclosed by the square and the circle?

200+25π200 + 25\pi

100+75π100 + 75\pi

75+100π75 + 100\pi

100+100π100 + 100\pi

100+125π100 + 125\pi

Solution:

The square has area 102=10010^2 = 100 and the circle has area π(10)2=100π.\pi(10)^2 = 100\pi.

Since the circle is centered at a vertex of the square, exactly one quarter of the circle, area 25π,25\pi, lies inside the square.

The union has area 100+100π25π=100+75π.100 + 100\pi - 25\pi = 100 + 75\pi.

Thus, the correct answer is B.

Problem 9 in Other Years