Geometry question

Hint: make the small length of the shaded area, ( the smaller side of the parallelogram x). Then you can use simular triangles.

Let the distance on the base line, from the right edge to the strip, be x.

Then the following applies:

$A_{rectangle}=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ A_{strip}=1\cdot \sqrt{8.5375^2+8^2}\\ \color{blue}A_{strip}=11.700$

!

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