2003 AMC 10A Problems/Problem 17

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

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

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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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