Difference between revisions of "2003 AMC 10A Problems/Problem 17"
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== Problem == | == 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? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math> \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 </math> | <math> \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 </math> | ||
Revision as of 18:11, 11 November 2015
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?
Solution
Let 
be the length of a side of the equilateral triangle and let 
be the radius of the circle.
In a circle with a radius the side of an inscribed equilateral triangle is 
.
So 
.
The perimeter of the triangle is 
The area of the circle is 
So:
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
