What is the relationship between an equilateral triangle's base and the radius of it's circumscribed circle?

Question

I know the length of the radius of the circle, and I want to know the length of the base of the triangle

Answer 1

From the law of cosines,
L^2 = R^2 + R^2 -2RRcos(π/3)
L^2 = 2R^2 -2R^2cos(π/3)
L^2 = 2R^2 + R^2 = 3R^2
L = R√3

Answer 2

you purely requested about the relation so i'm not giving the operating. The relation is L = sqrt(3) x D the position L is the length of the triangle D is the diameter of the circle. that is suggested as follows: The length of the area of the triangle is an similar as root 3 cases the diameter. desire you may understand this.

Answer 3

the circumradius is the hypotenuse of a right triangle with one leg of length equal to the length of the inradius and other leg of length equal to half the length of a side of the equilateral triangle. since an angle of an equilateral triangle is 60 degrees, and the angle bisectors coincide with the perpendicular bisectors, then:

letting R = length of circumradius
r = length of inradius
S/2 = other leg of the right triangle referred to above, where S is the length of a side of the equilateral triangle

R^2 = r^2 + (S/2)^2

but R = 2r, or R/2 = r

R^2 = (R/2)^2 + (S/2)^2
R^2 = (R^2)/4 + (S^2)/4
(3R^2)/4 = (S^2)/4
3R^2 = S^2
S = R*sqrt(3), where sqrt(3) is the positive square root of 3.

Answer 4

From the centre of the circle, draw a radius to each of two vertices of the triangle.
The smaller isosceles triangle formed now has sides r, r and x (where x is a side of the equilateral triangle), and angles 30º, 30º and 120º.
Then we can say that : x / sin(120º) = r / sin(30º).
Thus, x = r * sin(120º) / sin(30º)
= r * sin(60º) / sin(30º)
= r * [sqrt(3) / 2] / (1 / 2)
= r * sqrt(3)
or, base = radius * sqrt(3)

Answer 5

Here's another one
http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII35.html

so draw a straight line through an apex of your triangle bisecting the opposite edge and the circumscribing circle - this is a diameter of course. Choose the edge that you have just bisected as your other chord, now you have (base/2)^2 = height*(diameter-height)
and you can rearrange to find the diameter.
(base^2/4+height^2)/height= diameter

Answer 6

Figure out the diameter of the circle and you'll have the base of your equilateral.

Answer 7

let the side of the equilateral triangle be 'a'
and the radius of the circumcircle 'r'
a/2/r=cos30*
=>a/2r=rt3/2
multiplying by 2r
a=(rt3/2)(2r)=rt3r

aliter
a/sinA=2r
a/sin60*=2r
a=2rsin60*
=2r*rt3/2
=rt3r