2007 AMC 10A Problem 14 — Solution | LIVE by Po-Shen Loh

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2007 AMC 10A Problem 14

Below is the professionally curated solution for Problem 14 of the 2007 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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Concepts:inscribed angle right triangle triangle area

Difficulty rating: 1420

14.

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

$8.64$

$12$

$5\pi$

$17.28$

$18$

Solution:

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

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

The area is $\tfrac12 (3x)(4x) = 6x^2 = 6 \cdot \tfrac{36}{25} = 8.64.$

Thus, the correct answer is A.

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