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

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

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

Video Solution

https://youtu.be/El9eQJmDO78

~savannahsolver

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2003 AMC 10A Problems/Problem 17

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Problem

Solution

Video Solution

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