Perimeter of a triangle inscribed in a circle

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SUMMARY

The perimeter of an equilateral triangle inscribed in a circle can be calculated using the radius \( r \) of the circle. The formula for the perimeter \( P \) is \( P = 3 \times (r \times \sqrt{3}) \), which derives from the relationship between the radius and the side length of the triangle. The center of the circumscribed circle is the intersection of the triangle's perpendicular bisectors, confirming the geometric properties involved. This approach simplifies the problem by leveraging known geometric relationships.

PREREQUISITES
  • Understanding of basic geometry concepts, particularly triangles and circles.
  • Familiarity with the properties of equilateral triangles.
  • Knowledge of the circumcircle and its relationship to triangle vertices.
  • Ability to apply the Pythagorean theorem in geometric contexts.
NEXT STEPS
  • Research the derivation of the formula for the perimeter of triangles inscribed in circles.
  • Learn about the properties of circumcircles and their significance in triangle geometry.
  • Explore the relationship between the radius of a circumcircle and the side lengths of different types of triangles.
  • Study advanced geometric constructions involving perpendicular bisectors and their applications.
USEFUL FOR

Students studying geometry, educators teaching triangle properties, and anyone interested in the mathematical relationships between circles and polygons.

Fedcer
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Homework Statement


Find the Perimeter of an equilateral triangle inscribed in a circle knowing the radius r.

Homework Equations


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The Attempt at a Solution


Browsing the web I found that the intersection of the three perpendicular bisectors of a traingle is the center of it's circumscribed circle. Knowing this I proceeded as following: [img=http://img214.imageshack.us/img214/2971/problem1bgc6.th.jpg]
I think that what I did is correct. However, since I should've been able to solve it with what I already knew I thought that maybe there is another easier solution.

Thanks.
 

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Triangle AOB is isoceles. Just drawing the perpendicular from O to AB get you to the same position as the result you quote ( in fact it demonstrates it)
 

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