Difference between revisions of "2007 AMC 10B Problems/Problem 11"
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− | == Problem == | + | ==Problem== |
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− | <math>\ | + | 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? |
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+ | <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> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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</cmath> | </cmath> | ||
− | Substituting and solving gives <math>r = \frac {9}{4\sqrt {2}}</math>. Then the area of the circle is <math>r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \frac {81}{32} \pi | + | Substituting and solving gives <math>r = \frac {9}{4\sqrt {2}}</math>. Then the area of the circle is <math>r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}} | R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}} | ||
</cmath> | </cmath> | ||
− | and the answer is <math>R^2 \pi = \mathrm{(C)}</math> | + | and the answer is <math>R^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math> |
Alternatively, by the Extended [[Law of Sines]], | Alternatively, by the Extended [[Law of Sines]], |
Revision as of 15:30, 5 June 2011
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 |