Find the perimeter of a polygon $ABCDEF$

A circle, with a radius of $12$ cm and with the center coinciding with the center of an equilateral triangle with a side of $36$ cm, intersects the sides of the triangle at points $A, B, C, D, E$ and $F$. Find the perimeter of the polygon $ABCDEF$.

Image of the question:

I solved it as follows: I proved that this polygon is a regular hexagon, found its side, it is $12$ and then the perimeter is $6⋅12=72$. I was confused by the simplicity of the task. Did I solve the problem correctly?

Solution 1:

Let G and H, I be the midpoints of BC and DE, as O is the center OG, OH, OI are perpendicular to BC, DE, AF and let R be the radius of the circumscribed circle of the equilateral triangle,

You can find OG and OH by pyth. theorem,

$OG=\sqrt{R^2-18^2}$

$OH=\sqrt{R^2-18^2}$

$OI=\sqrt{R^2-18^2}$

by these

$OG=OH=OI=h$

Also by pyth. theorem you can get the lengths of BY, CZ, DZ, XE, XF, AY

$BY=18-\sqrt{12^2-h^2}$

$CZ=18-\sqrt{12^2-h^2}$

$DZ=18-\sqrt{12^2-h^2}$

$XE=18-\sqrt{12^2-h^2}$

$XF=18-\sqrt{12^2-h^2}$

$AY=18-\sqrt{12^2-h^2}$

With above you can find out $\triangle ABY, \triangle DCZ, \triangle XFE$ are equilateral and the side lengths of each other are equal.

By angle chasing you can find that the all angles of the hexagon is equal 120 and it is a regular hexagon.

As OB bisects $\angle ABC$, $\angle ABO=60$ and the one side of the hexagon will be equal to 12 (radius of the inscribed circle)

And the perimeter will be equal to = 12*6=72

Your answer is correct here is the proof for it