2007 AMC 10B Problem 11

Below is the professionally curated solution for Problem 11 of the 2007 AMC 10B, 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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Concepts:circumcircle, circumcenter, and circumradius triangle area circle area

Difficulty rating: 1460

11.

A circle passes through the three vertices of an isosceles triangle that has two sides of length $3$ and a base of length $2.$ What is the area of this circle?

$2\pi$

$\dfrac{5}{2}\pi$

$\dfrac{81}{32}\pi$

$3\pi$

$\dfrac{7}{2}\pi$

Solution:

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

The circumradius is $R=\dfrac{abc}{4K}=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2}=\dfrac{9}{4\sqrt2} =\dfrac{9\sqrt2}{8}.$

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

Thus, the correct answer is C.

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