+0 A rectangle contains a strip of width $1,$ as shown below. Find the area of the strip.
0 Hint: make the small length of the shaded area, ( the smaller side of the parallelogram x). Then you can use simular triangles.
+14943 Let the distance on the base line, from the right edge to the strip, be x.
Then the following applies:
\( A_{rectangle}=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0 \)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ A_{strip}=1\cdot \sqrt{8.5375^2+8^2}\\ \color{blue}A_{strip}=11.700\)
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