Equilateral triangle with side 8 inches is inscribed in a circle. What is the radius of the circle?

Answer 1

One thing you can do is note that the distance from the center of the equilateral triangle to any vertex of the triangle will be the radius of the circle.

The first thing to note is that a full circle is 360 degrees, so if we draw a line from the center to each of the vertex points, the angle will effectively be split into 3, with each angle being 360/3 degrees. That's 120 degrees.

Now, note that each triangle created just now is an isoceles triangle, so the other two angles would be equal. Since the sum of the angles of a triangle is 180 degrees, and one angle is already 120, that would mean the other two angles would be 30 and 30.

So, right now we have a triangle with angles A=120,B=30,C= 30 with opposite side a = 8.

By the Sine law,

8/[sin(120)] = r/sin(30)

Therefore, r = 8*[sin(30)/sin(120)]

The exact values of sin(30) and sin(120) are actually known.
sin(30) = 1/2, and sin(120) = sin(2pi/3) = sqrt(3)/2

Therefore, r = 8*[1/sqrt(3)] = 8/sqrt(3)

Answer 2

First, recall that the ratio of short leg:long leg:hypotenuse is 1:sqrt(3):2.

Now draw a small 30-60-90 triangle within the larger triangle so that the long leg is 1/2 of a side of the equilateral triangle. The short leg will be perpendicular to this side. Thus, the hypotenuse will go from a vertex of the triangle to the center of the circle. (You can draw six such triangles in an equilateral triangle, but we only need to consider one.)

Now, since the long leg of our small triangle is 4. The ratio of hypotenuse:long leg is 2:sqrt(3). So algebra tells us the hypotenuse length:

hypotenuse:long leg is 2:sqrt(3)
hypotenuse:4 = 2/sqrt(3)
hypotenuse = 8/sqrt(3)

The hypotenuse of this triangle is also the radius of the circle.

Answer 3

I'm not entirely sure on this, as I've forgotten some of my Geometry, so don't rely on this unless other people post similar results....

To find the radius of the circle, you must find the point in the triangle in which all distances from each vertex is equal. This is always where the perpendicular bisectors to all 3 sides intersect. I can't remember if there's an easier way to do this next part, but it works :-/

Next, you must find the distance from this point to a vertice. To do this, I place the triangle on a coordinate plane, with one vertice at 0. y = 0, y=sqrt(3)x, and y=-sqrt(3)x+sqrt(192) are the lines I calculated to make the triangle. Now, using the midpoint formula, Just find the intersection of 2 lines (The third will intersect at the same point). Midpoint formula ((xsub2 + xsub1)/2,(ysub2+ysub1)/2)) shows that the midpoint for (0,0) and (0,8) (The points given by the line y=0) is (0,4). That was the easy one. For the second, I'm using (0,0) and (4,sqrt(48)) as my points. (4,sqrt(48)) is the point exactly 8 units from the origin at a 60 degree angled line. That gives us (2,sqrt(12)). Now, we need to find a line perpendicular to our other lines that pass through those points. The first one is easy, for (4,0). x = 4 will do. Now, perpendicular lines have opposite recipricals as their slope. The opposite reciprical of sqrt(3) is negative sqrt(1/3). So, sqrt(12) = -sqrt(1/3)(2) +b. Multiply -sqrt(1/3) by 2 and subtract from both sides, and you get sqrt(64/3) = b. So, y=-sqrt(1/3)x+sqrt(64/3) and x=4. Solve for these lines by filling in x for 4 on the complicated one, and you get y as sqrt(16/3). So, (sqrt(16/3),4) is your middle. Distance formula (Distance = sqrt((xSub2-xSub1)^2 + (ySub2-ySub1)^2). Filling in for out center and the origin, we get our distance to sqrt(64/3), or roughly 4.6188021535170611.

Sorry for the previous wrong answer, I forgot a negative :-/ Tis better now, and right =)

Answer 4

Drop a perpendicular from each vertex of the triangle to the base of the opposite side. This bisects each vertex and also the opposite side. The intersection of these perpendiculars creates isosceles triangles on each side whose legs are radii of the circle. Each of these isosceles triangles is further divided into two congruent 30-60-90 degree triangles by the perpendicular dropped from the opposite vertex (you figure out why). 30-60-90 degree triangles have special properties. We recall that the cosine of 30 degrees is (sq rt 3)/2 and note that a radius of the circle is now the hypotenuse of each 30-60-90 degree triangle.

Now we can set up this equation:

[(sq rt 3)/2] = 4/x.

Performing the algebra, x = [8*(sq rt 3)]/3 inches.

If we divide this out, x is approximately equal to 4.61 inches.

Answer 5

Or, you can use the formula for the length of a chord instead of deriving it:

c = 2r sin(theta/2)

where theta is the central angle distended by the chord, in this case 120.

8 = 2 * r * sin(60)
8 = 2 * r * sqrt(3)/2
r = 8 / sqrt(3) = 8sqrt(3) / 3

Answer 6

in equilaterial triangle R = 2/3 a * square root of 3 /2 = 8 / square root of 3