A triangle's rotational kinetic energy

[aitee]

Homework Statement

The three 280 g masses in the figure are connected by massless, rigid rods to form a triangle.

What is the triangle's rotational kinetic energy if it rotates at 6.5 revolutions per s about an axis through the center?

Values given:
Mass of each ball at the corner of the triangle: 0.3 kg
Sides of Equilateral Triangle: 0.4m

Homework Equations

I= 3*(mr^2)
KE =.5 Iw^2

The Attempt at a Solution

I know that there is another post for this same question, but that user figured it out and didn't post their error in thinking and I'm stuck in the same rut.

I first tried to solve it like this:

H = sin(60) * .2
I then attempted to use half of this height to find the 'r' value. The r value bisects the 60 degrees, hence 30 degrees.

So r = 1/(sin(30)/.5H)

I know this is wrong, but I don't know how to go about solving this. (is the center point even centered relative to the top point to the middle intersection of the lower rod?)

I searched online and found someone who gave this as a solution to finding r:
r = (2/3)*median = (2/3)[sq rt(0.40^2 - 0.20^2)] = 0.231 m
which when I attempted this it worked. But I'm not familiar with this equation and wanted to understand how this works, and if there was another way (as I'm thinking our teacher expected us to use)?

Help in understanding this would be so greatly appreciated.

 

[issacnewton]

well in that formula you have put, they are just using Pythagoras theorem. side of the triangle is the hypotenuse, line joining the vertex and the midpoint of the line in front of the
vertex is one side of this right angle triangle