Solving Pythagoras Problem An aircraft hangar is semi-cylindrical, with diameter 40m and length 50 m. A helicopter places an inelastic rope across the top of the hangar and one end is pinned to a corner, a A. The rope is then pulled tight and pinned at the opposite corner, B. Determine the lenghth of the rope.
So, first I find the diagonal line from A straight to B. 
c^2=50^2+40^2
The answer is 64.03124......
Then, I find out that there is a semi-circle shape. So I find the length of the arc.
r=32.05162.......
Length=2*Pi*32.05162.....
      =201.16008m......  
But the correct answer is 80.30m
Can anyone tell me where i did wrong?


The rope does not form a semicircle. Rather it is an ellipse. So a simple semicircle calculation will be wrong.
The best suggestion above is to flatten out the roof of the hangar. Imagine just lifting the metal off and squashing it flat. You have a rectangle. Length = length of hangar. Width = half the circumference of the circle at the front of the hangar.
Then use Pythagoras to find the length of the rope, as diagonal of that rectangle.
