Inspired from a recent post: How does one obtain the radius

Question

Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?

young_gun · Answer

the answer is x sqrt(3)/3
actual solutions, as opposed to answers, would be greatly appreciated  .

walker · Answer

for me: \frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2} r=\frac{1}{sqrt{3}}x

young_gun · Answer

any non-trig way to do this?

walker · Answer

r^2=(\frac{x}{2})^2+(\frac{r}{2})^2

r=\frac{1}{sqrt{3}}x

young_gun · Answer

thanks, but do you think you could post a diagram or explain more...i'm not seeing it. seems like you used pythag? where does (r/2)^2 come from? is it known that in an equil. triangle, the mid point to the middle of the sides is r/2?

walker · Answer

hope this help

GMATBLACKBELT · Answer

knew that question would spark some interest. Walker has a nice picture, but essentially the long side of the smaller triangle is x/2 then small side is xsqrt3/6 So the hypt. is xsqrt3/3 Which is the radius. or as walker puts it x/sqrt3

testgmat · Answer

Great analysis of the question. You guys deserve some kudos.

Inspired from a recent post: How does one obtain the radius

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