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 square of the ratio of their side lengths is equal to the ratio of their areas. Call the area of the triangle 
.
2+3\sqrt{2})^2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
