Difference between revisions of "2003 AMC 10A Problems/Problem 17"
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Let <math>s</math> be the length of a side of the equilateral triangle and let <math>r</math> be the radius of the circle. | Let <math>s</math> be the length of a side of the equilateral triangle and let <math>r</math> be the radius of the circle. | ||
− | In a circle with a radius <math>r</math> the side of an inscribed equilateral triangle is <math>r\sqrt{3}</math>. | + | In a circle with a radius <math>r</math>, the side of an inscribed equilateral triangle is <math>r\sqrt{3}</math>. |
So <math>s=r\sqrt{3}</math>. | So <math>s=r\sqrt{3}</math>. |
Latest revision as of 09:28, 1 July 2017
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 |
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