Hey Denis, hope you had a nice Christmas.
\(\displaystyle 2sqrt{11}(10)=66.332\) is the area of an isosceles.
Also, \(\displaystyle \frac{1}{2}(144)sin(1.97022156668)=66.332\) is the area of an isosceles.
This is just another isosceles area formula besides the one you used. I found the derivation in an old trig/alg book. The image is below. That's easier tham typing the LaTex.
First, I solved for theta in the chord formula, which can be seen by my previous post. theta is 1.97022156668 radians or 112.88538 degrees.
The formula for a segment, what I have shaded in my haphazard diagram, is given by \(\displaystyle \frac{1}{2}r^{2}({\theta}-sin({\theta}))\). It just takes the sector area and subtracts the triangle area, giving segment area.
Which gives:\(\displaystyle \frac{1}{2}(144)(1.97022156668-sin(1.97022156668))=75.523456994 m^{2}\)
We can check our answers by finding the middle ordinate, \(\displaystyle r(1-cos(\frac{\theta}{2}))=12(1-cos(.98511078334))=5.36675041931=>12-5.36675041931=2\sqrt{11}\)
Calc:
\(\displaystyle 2\int_{-12}^{-2sqrt{11}}sqrt{144-x^{2}}dx=75.5234569935\)
Sorry, if I got carried away. It's just fun to check different methods to see if it comes out the same. It did.