document.write( "Question 612858: What is the diameter of a circle with an inscribed equilateral triangle with side lengths of 6? \n" ); document.write( "

Algebra.Com's Answer #385794 by KMST(5328) $\"\"$ $\"About$

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All 3 angles in triangle ABC measure $\"60%5Eo\"$ ,AB=BC=AC=6, and AO=BO=CO=radius of circle.
\n" ); document.write( "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.
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AB=BC=AC=6, AD=BD=3, AO=BO=CO=radius of circle,
\n" ); document.write( "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.
\n" ); document.write( "Pythagoras theorem says that
\n" ); document.write( " $\"AC%5E2=AD%5E2%2BCD%5E2\"$ so, substituting,
\n" ); document.write( " $\"6%5E2=3%5E2%2BCD%5E2\"$ --> $\"36=9%2BCD%5E2\"$ --> $\"36-9=CD%5E2\"$ --> $\"CD%5E2=27\"$
\n" ); document.write( "So $\"CD=sqrt%2827%29=sqrt%289%2A3%29=sqrt%289%29%2Asqrt%283%29=3sqrt%283%29\"$
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\n" ); document.write( "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.
\n" ); document.write( "In all those 30-60-90 triangles, the hypotenuse is twice as long as the short leg:
\n" ); document.write( " $\"AO%2FDO=AC%2FAD=6%2F3=2\"$ so $\"AO=2DO\"$ , $\"CD=CO%2BDO=AO%2BDO=2DO%2BDO=3DO\"$ <--> $\"DO=%281%2F3%29CD\"$
\n" ); document.write( "So $\"CO=CD-DO=CD-%281%2F3%29CD=%281-1%2F3%29CD=%282%2F3%29CD=%282%2F3%293sqrt%283%29=2sqrt%283%29\"$ \n" ); document.write( "

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