SOLUTION: What is the diameter of a circle with an inscribed equilateral triangle with side lengths of 6?

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Question 612858: What is the diameter of a circle with an inscribed equilateral triangle with side lengths of 6?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
All 3 angles in triangle ABC measure 60%5Eo,AB=BC=AC=6, and AO=BO=CO=radius of circle.
So triangles ABO, BCO, and ACO are isosceles and congruent. Their base angles measure 30%5Eo, and their altitudes are perpendicular to the base and bisect the base. So we can draw DO, the altitude of ABO. DO is part of DC, the altitude of ABC, which separates congruent right triangles ACD and BCD.
AB=BC=AC=6, AD=BD=3, AO=BO=CO=radius of circle,
Triangles ACD and BCD are congruent 30-60-90 right triangles. They have a 30%5Eo angle at C, a 60%5Eo angle, and a 90%5Eo angle.
Pythagoras theorem says that
AC%5E2=AD%5E2%2BCD%5E2 so, substituting,
6%5E2=3%5E2%2BCD%5E2 --> 36=9%2BCD%5E2 --> 36-9=CD%5E2 --> CD%5E2=27
So CD=sqrt%2827%29=sqrt%289%2A3%29=sqrt%289%29%2Asqrt%283%29=3sqrt%283%29

DO splits triangle ABO (isosceles triangle with 30%5Eo base angles) into two congruent 30-60-90 right triangles: AOD and BOD. They are similar to triangles ACD and BCD.
In all those 30-60-90 triangles, the hypotenuse is twice as long as the short leg:
AO%2FDO=AC%2FAD=6%2F3=2 so AO=2DO, CD=CO%2BDO=AO%2BDO=2DO%2BDO=3DO <--> DO=%281%2F3%29CD
So CO=CD-DO=CD-%281%2F3%29CD=%281-1%2F3%29CD=%282%2F3%29CD=%282%2F3%293sqrt%283%29=2sqrt%283%29