How can I figure out the radius of a circle, when all i have to work with is an equilateral triangle?

Question

Ok so i have to make a icosahedron (a 20 sided polygon) tonight, and i need the triangles to be 6 inches on each side (equilateral triangles).
i know HOW to make the triangle..once i have the circle. I just don't know how big to make the circle!!
SO my question:
How can I figure out the radius of the circle using only a triangle of 6 inches on each side? or, What would the radius of the circle be?! (THAT would be the mooost helpful answer)

OK IM SORRY I DIDNT KNOW THAT IT WAS A POLYHEDRON. GOT IT.

i didnt understand anything anyone told me. im only in GEOMETRY and this is for a chemistry project..i dont know how to do trig yet!!

Answer 1

What circle?

Also, an icosahedron isn't a polygon, it's a polyhedron (3-D)

Just make the 20 triangles and attach them to form a ball-like shape.

http://upload.wikimedia.org/wikipedia/en/e/e2/Icosahedron.gif

Answer 2

An icosahedron is not a 20-sided polygon it is one of 5 Platonic solids. A twenty sided polygon is an icosagon and it would be made up of 20 isosceles triangles (2 equal sides).

There is a formula for determining the radius of a sphere which is determined by the icosahedron and you can work it out with trig. Actually there are two possible spheres associated with an icosahedron, one circumscribed (touching all the vertices) and an inscribed sphere (one tangent to the center of the 20 equilateral triangles.

When you said 20 sided polygon I am assuming you meant a 20 sided polyhedron.

One of these the icosagon is a two dimensional figure and the icosahedron is a solid figure. To me it makes no sense to try to associate an equilateral triangle with the two dimensional figure.

The 20 6 inch equilateral triangles will form an icosahedron but working out the radii of the two associated spheres seems straight forward but maybe a little lengthy. Good luck. I'll star your question and look later.

Answer 3

An equilateral triangle will fit inside a circle that touches each point of the triangle. To find the center of the triangle, and the radius of the circle, draw a right triangle by bisecting one leg of the triangle and draw a perpendicular line toward the center of the triangle. Then bisect the 60 degree angle at the point of the traingle, and draw a line toward the center of the triangle at 30 degrees until it meets the first line. Now you have a 30/60/90 degree right angle, with a base/adjacent side of 3" and an angle of 30 degrees. Solve:

Cosign of 30 degrees=3"/hypotenuse

.867=3/h

.867xh=3/hxh

.867xh=3

.867xh/.867=3/.867

h=3.46