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?

 A: Hint: imagine the semicylindrical surface of the hangar as planar and apply Pytagora's theorem. If you flatten it to get a rectangle, the curve that you search becomes the diagonal of that rectangle.
A: You could Construct a Right angle triangle with following dimensions
$$h=50$$
(Since It's Length of Cylinder)
$$b=\pi\times20$$
(Since $20$ is Radius and If I'm not wrong Circumference of Semicircle would be $\pi r$)
so Length of rope will be
$$L=\sqrt{h^2+b^2}$$
$$L=\sqrt{2500+400\pi^2}$$
$$L\approx80.30$$
Approximated Here
A: This question is a simple Pythagoras problem.
 First, we find the length of the arc whose diameter is 40m.
Then we assume it to be a side of the quadrilateral whose side is 50 m.
Next, we use Pythagoras theorem to find the diagonal.
The answer will be near to 80.34..
A: 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.
