2007 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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Concepts:right trianglecircumcircle, circumcenter, and circumradiustriangle area

Difficulty rating: 1290

10.

A triangle with side lengths in the ratio 3:4:53:4:5 is inscribed in a circle of radius 3.3. What is the area of the triangle?

8.648.64

1212

5π5\pi

17.2817.28

1818

Solution:

Let the sides be 3x,3x, 4x,4x, and 5x.5x. The triangle is right, so its hypotenuse is a diameter.

Thus 5x=23=6,5x=2\cdot 3=6, giving x=65.x=\tfrac65.

The area is 123x4x=6x2=63625=21625=8.64.\tfrac12\cdot 3x\cdot 4x=6x^2=6\cdot\tfrac{36}{25}=\tfrac{216}{25}=8.64.

Thus, the correct answer is A.

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