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? !An aircraft hangar

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&oq=sqrt\(2500%2B400*pi*pi\)&aqs=chrome..69i57.23046j0j7&sourceid=chrome&es_sm=93&ie=UTF-8)