2007 AMC 10B Problem 11

Problem 11 of 25IntermediateGeometry

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?

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Solution

The triangle has sides

3,

3,

and

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.