2007 AMC 10B Problem 11

Question

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

Use the circumradius formula

R=\dfrac{abc}{4K},

where

K

is the triangle’s area

The sides are

3,

3,

and

2;

find the area from the base and height, then compute

R

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.

Problem 11 in other years