2007 AMC 10B Problem 11

Attempt Problem 11 of the 2007 AMC 10B 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 2007 AMC 10B solutions, or check the answer key.

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11.

A circle passes through the three vertices of an isosceles triangle that has two sides of length 33 and a base of length 2.2. What is the area of this circle?

2π2\pi

52π\dfrac{5}{2}\pi

8132π\dfrac{81}{32}\pi

3π3\pi

72π\dfrac{7}{2}\pi

Answer: C
Concepts:circumcircle, circumcenter, and circumradiustriangle areacircle area
Difficulty rating: 1460
Solution:

The triangle has sides 3,3,2.3,3,2. Its area is 1223212=22.\dfrac12\cdot 2\cdot\sqrt{3^2-1^2}=2\sqrt2.

The circumradius is R=abc4KR=\dfrac{abc}{4K} =332422=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2} =942=\dfrac{9}{4\sqrt2} =928.=\dfrac{9\sqrt2}{8}.

The area of the circle is πR2=π81264=8132π.\pi R^2=\pi\cdot\dfrac{81\cdot 2}{64}=\dfrac{81}{32}\pi.

Thus, the correct answer is C.

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