Solution: Original radius $ r = \sqrt\frac100\pi\pi = 10 \, \textkm $. New radius $ r' = 8 \, \textkm $. Original area $ 100\pi $, new area $ \pi (8)^2 = 64\pi $. The decrease is $ 100\pi - 64\pi = \boxed36\pi \, \textkm^2 $. - Groen Casting