Difference between revisions of "2007 AMC 12A Problems/Problem 10"

Revision as of 08:14, 19 October 2007

Problem

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

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 $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$

2007 AMC 12A (Problems • Answer Key • Resources)
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
All AMC 12 Problems and Solutions

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 [[Image:2007_AMC12A-10.png]]
-Since 3-4-5 is a [[Pythagorean triple]], the triangle is a [[right triangle]]. Since right triangles can be inscribed in [[semicircle]]s, the hypotenuse is <math>2r = 6</math>. Then the other legs are <math>\frac{24}5=4.8</math> and <math>\frac{18}5=3.6</math>. The area is <math>\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}</math>
+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 <math>2r = 6</math>. Then the other legs are <math>\frac{24}5=4.8</math> and <math>\frac{18}5=3.6</math>. The area is <math>\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}</math>
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Difference between revisions of "2007 AMC 12A Problems/Problem 10"

Revision as of 08:14, 19 October 2007

Problem

Solution

See also