+729 A rectangle contains a strip of width $1,$ as shown below. Find the area of the strip.
+1365 Let x be the length from the right edge to the strip.
We have
\([rectangle]=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ [strip]=1\cdot \sqrt{8.5375^2+8^2}\\ [strip]=11.7\)
Thanks! :)
