+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.
+15001 Let the distance on the base line, from the right edge to the strip, be x.
Then the following applies:
\( A_{rectangle}=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0 \)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ A_{strip}=1\cdot \sqrt{8.5375^2+8^2}\\ \color{blue}A_{strip}=11.700\)
!
+129850 
Let AE = "a"
Note that angle AGE = 90 because GE is perpendicular to AF
Let angle GAE = theta
Because AF is parallel to EC , then angle GAE = CED
sin ( GAE) = 1/a
tan (CED) = 8 / (10 - a)
sin (CED) = 8 / sqrt [ 8^2 + (10-a)^2] = 8 /sqrt [ a^2 -20a + 164]
sin (GAE) = sin (CED)
1/a = 8 / sqrt [a^2 -20a + 164 ] square both sides and cross-multiply
a^2 -20a + 164 = 64a^2
63a^2 + 20a - 164 = 0
a ≈ 1.4625
The strip is a parallelogram with a height of 8 and a width of a = 1.4625
Area = 8 * 1.4625 = 11.7
Note something interesting.......the numerical value of the area ≈ the numerical value of the length of AF, EC
