Problem:
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?
Answer Choices:
A. 2Ο
B. 25βΟ
C. 3281βΟ
D. 3Ο
E. 27βΟ
Solution:
Let BD be an altitude of the isosceles β³ABC, and let O denote the center of the circle with radius r that passes through A,B, and C, as shown.
Then
BD=32β12β=22β and OD=22ββr
Since β³ADO is a right triangle, we have
r2=12+(22ββr)2=1+8β42βr+r2, and r=42β9β=89β2β
