AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
Personal tools

2007 AMC 12A Problems/Problem 10

From AoPSWiki

The following problem is from both the 2007 AMC 12A #10 and 2007 AMC 10A #14, so both problems redirect to this page.

Problem

A triangle with side lengths in the ratio 3 : 4 : 5 is inscribed in a circle with radius 3. What is the area of the triangle?

\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18

Solution

Image:2007_AMC12A-10.png

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 2r = 6. Then the other legs are \frac{24}5=4.8 and \frac{18}5=3.6. The area is \frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}

See also

2007 AMC 12A (ProblemsResources)
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
2007 AMC 10A (ProblemsResources)
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us