Difference between revisions of "2003 AMC 10A Problems/Problem 17"

Revision as of 22:45, 16 July 2014

Problem

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

$\mathrm{(A) \ } \frac{3\sqrt{2}}{\pi}\qquad \mathrm{(B) \ } \frac{3\sqrt{3}}{\pi}\qquad \mathrm{(C) \ } \sqrt{3}\qquad \mathrm{(D) \ } \frac{6}{\pi}\qquad \mathrm{(E) \ } \sqrt{3}\pi$

Solution

Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle.

In a circle with a radius $r$ the side of an inscribed equilateral triangle is $r\sqrt{3}$ .

So $s=r\sqrt{3}$ .

The perimeter of the triangle is $3s=3r\sqrt{3}$

The area of the circle is $\pi r^{2}$

So: $\pi r^{2} = 3r\sqrt{3}$

$\pi r=3\sqrt{3}$

$r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(A)}\ \frac{3\sqrt{3}}{\pi}}$

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 <math>\pi r=3\sqrt{3}</math>
-<math>r=\frac{3\sqrt{3}}{\pi} \Rightarrow B</math>
+<math>r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(A)}\ \frac{3\sqrt{3}}{\pi}}</math>
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Difference between revisions of "2003 AMC 10A Problems/Problem 17"

Revision as of 22:45, 16 July 2014

Problem

Solution

See Also