Solution: Original radius $ r = \sqrt{\frac{100\pi}{\pi}} = 10 \, \text{km} $. New radius $ r' = 8 \, \text{km} $. Original area $ 100\pi $, new area $ \pi (8)^2 = 64\pi $. The decrease is $ 100\pi - 64\pi = \boxed{36\pi} \, \text{km}^2 $.