Printable

2007 AMC 10A Problem 14

2007 AMC 10A Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 2 of 5 · about 2 min · position P11-15

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?

A)
$8.64$
B)
$12$
C)
$5\pi$
D)
$17.28$
E)
$18$

Answer

Key insight

A 3-4-5 triangle is right, so its hypotenuse is a diameter: 6 = (6/5)*5, and area scales by (6/5)^2 from the 3-4-5 area of 6.

Solution

Since $3^2 + 4^2 = 5^2$ , the triangle is a right triangle with the side of ratio $5$ as hypotenuse. A right angle inscribed in a circle subtends a diameter, so that hypotenuse is a diameter of the circle: length $2 \cdot 3 = 6$ .

The triangle is therefore the $3$ - $4$ - $5$ triangle scaled by $\frac{6}{5}$ . The $3$ - $4$ - $5$ triangle has area $\frac12 \cdot 3 \cdot 4 = 6$ , and areas scale by the square of the length ratio:
$6 \cdot \left(\frac{6}{5}\right)^2 = 6 \cdot \frac{36}{25} = \frac{216}{25} = 8.64.$

The answer is $\boxed{\textbf{(A)}\ 8.64}$ .

Why this works

Two facts do all the work: recognizing the Pythagorean triple, and Thales' theorem that the hypotenuse of an inscribed right triangle is a diameter. After that, a scale factor converts a known area; there is no need to compute the legs $3.6$ and $4.8$ separately, although doing so gives the same $\frac12 \cdot 3.6 \cdot 4.8 = 8.64$ .

The trap

Using the radius 3 as the hypotenuse instead of the diameter 6, which gives area 2.16, or answering 6 for the unscaled triangle.

Common mistakes

Using the radius $3$ as the hypotenuse instead of the diameter $6$ , which gives area $2.16$ , or answering $6$ for the unscaled triangle.
Scaling the area by $\frac65$ instead of $\left(\frac65\right)^2$ , giving $7.2$ .

Techniques

Set up the equation/formula and compute; no special trick needed