+8 When mineral deposits formed a coating 1 mm think on the inside of a pipe, the area through which which can flow was reduced by 20%. Find the original inside diameter of the pipe.
+118724 Hi NavySealFleetAdmiral Welcome to web2.0calc forum 
Your question looks interesting :)
When mineral deposits formed a coating 1 mm think on the inside of a pipe, the area through which which can flow was reduced by 20%. Find the original inside diameter of the pipe.
Let the original radius be r
\(Original \;Area =\pi r^2\\ Reduced\; Area =\pi(r-1)^2\\ 100\%-20\%=80\%\\ 80\%\;of\;\pi r^2=\pi(r-1)^2\\ 0.8*\pi r^2=\pi(r^2-2r+1)\\ 0.8 r^2=r^2-2r+1\\ 0.2r^2-2r+1=0\\ r^2-10r+5=0\\ r=\frac{10\pm\sqrt{100-20}}{2}\\ r=\frac{10\pm\sqrt{80}}{2}\\ r=\frac{10\pm 4\sqrt{5}}{2}\\ r=5\pm 2\sqrt{5}\\ r\approx 5\pm 4.47\\ BUT\;r>1\\ r\approx 9.47\;mm\\ \)
Check: 0.8*pi*9.47^2 = 225.3926852858567108
pi*8.47^2 = 225.3806844019199977
Yes close enough :)
