2007 AMC 12A Problems/Problem 10
- The following problem is from both the 2007 AMC 12A #10 and 2007 AMC 10A #14, so both problems redirect to this page.
Contents
Problem
A triangle with side lengths in the ratio is inscribed in a circle with radius 3. What is the area of the triangle?
Solution
Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is . Then the other legs are and . The area is
Solution 2
The hypotenuse of the triangle is a diameter of the circumcircle, so it has length . The triangle is similar to a 3-4-5 triangle with the ratio of their side lengths equal to . The area of a 3-4-5 triangle is . The square of the ratio of their side lengths is equal to the ratio of their areas. Call the area of the triangle . Therefore,
~mobius247
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 12 Problems and Solutions |
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
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