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

Revision as of 11:19, 15 January 2008

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

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

$r=\frac{3\sqrt{3}}{\pi} \Rightarrow B$

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

@@ Line 23: / Line 23: @@
 == See Also ==
-*[[2003 AMC 10A Problems]]
+{{AMC10 box|year=2003|ab=A|num-b=16|num-a=18}}
-*[[2003 AMC 10A Problems/Problem 16|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 18|Next Problem]]
 [[Category:Introductory Geometry Problems]]

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

Revision as of 11:19, 15 January 2008

Problem

Solution

See Also