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

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2007 AMC 10A Problem 14

Attempt Problem 14 of the 2007 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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?

$8.64$

$12$

$5\pi$

$17.28$

$18$

Answer: A

Concepts:inscribed angle right triangle triangle area

Difficulty rating: 1420

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 $\begin{aligned} \tfrac12 (3x)(4x) &= 6x^2 = 6 \cdot \tfrac{36}{25} \\ &= 8.64. \end{aligned}$

Thus, the correct answer is A.

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