kaps786 wrote:Hi Everyone - Many thanks for your replies, guess will have to mug up these centroid, radius etc formulas - have been not using them for a while, so out of touch with them.
Mitch - Thanks, particularly liked your simplistic solution, since it uses no formula as such , but could not understand why the line joining the centre of the circle to the equilateral triangle bisects it - i.e divides it into 30 and 30 to make that triangle 30-60-90.
That might be some property of triangles that I am missing out.
After one figures out its 30-60-90 then its certainly easier, but I could not notice that 30-60-90 triangle in there.
Thanks again to all for sharing your knowledge.
For your peace of mind, here's further proof of the 30-60-90 triangle:
OB and OC are radii, so OB = OC = r.
The intersection of a line tangent to a circle and a radius drawn to the point of tangency forms a right angle.
Thus, triangles ABO and ACO are right triangles.
Thus:
AB^2 + r^2 = AO^2
AC^2 + r^2 = AO^2.
Thus:
AB^2 + r^2 = AC^2 + r^2
AB^2 = AC^2
AB = AC.
Thus, triangles ABO and ACO are congruent, and each is a 30-60-90 triangle.
If we extend this logic, we'll get the following figure:
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