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RUSH - (exam on monday) Equal. Triangles in a Circle

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by Kiddo25 » Fri Apr 03, 2009 12:11 pm
If an equalilateral triangle is inscribed in a circle, with perimeter of the triangle = 24. What is the approximate value for the diameter of the circle?

so all we know is each side of the equal. triangle is 8, and cutting it in half makes 30-60-90 with:
t = 4
2t = 8 (each side of the original equal. triangle)
t sq root 3 = 4 sq root 3

How can I find the diameter?
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by ncr_10 » Sat Apr 04, 2009 1:03 am
You must note that the radius of a circumscribed circle is 2/3 * (height of the equilateral triangle).

So in this case the radius would be (2*(4 sqrt(3))) / 3 = (8/(sqrt(3))).

The shortcut to find radius would be a/sqrt(3), where a is the length of the side of the triangle.

-ncr
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by cbrombe1 » Tue Apr 14, 2009 11:08 am
Can you please clarify (maybe with a picture)? Thanks!
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by nasa » Tue Apr 14, 2009 9:18 pm
If an equalilateral triangle is inscribed in a circle, with perimeter of the triangle = 24. What is the approximate value for the diameter of the circle?

Call the triangle ABC with side length "a" and the center of circle O
The hieght h = a*sqrt3/2
perimeter = 24 ==> a = 8

Look at the triange OBC making an angle 120deg at the center. (Since angle BAC is 60deg, angle BOC is 120deg) So ang OAB = (180-120)/2 = 30deg

Also, you can consider the new hight hn = h/2, due to the fact ABC is a equilateral triangle.
hn/R = sin30
==> diameter = 2R = 2*hn/sin30 = 2*h/(2*sin30) = h = 8 * 1.732 = 13.8 approx. value
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