Difference between revisions of "2007 AMC 10B Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | A circle passes through the three vertices of an isosceles triangle that has two sides of length <math>3</math> and a base of length <math>2 | + | A circle passes through the three vertices of an isosceles triangle that has two sides of length <math>3</math> and a base of length <math>2</math>. What is the area of this circle? |
<math>\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi</math> | <math>\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi</math> |
Revision as of 19:42, 5 June 2011
Contents
[hide]Problem
A circle passes through the three vertices of an isosceles triangle that has two sides of length and a base of length . What is the area of this circle?
Solution
Solution 1
Let have vertex and center , with foot of altitude from at .
Then by Pythagorean Theorem (with radius , height ) on
Substituting and solving gives . Then the area of the circle is .
Solution 2
By (or we could use and Heron's formula), and the answer is
Alternatively, by the Extended Law of Sines, Answer follows as above.
See also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |