Folded rectangle

In the modified image above, the blue triangles have the same area, and the two red triangles also have the same area. The rectangle consists of two blue and two red triangle, whereas the pentagon consists of two blue and only one red triangle. If we were able to calculate the  area of each colored triangle, we would be able to easily calculate the area of the pentagon. So here is how we could find the areas we need.

The blue tringle FBC is a right triangle with length of the hypothenuse labeled y and  legs of length 4 and length x, as indicated. we can write the following two equations involving x and y:

\(x+y=10\)

\({y}^{2}={x}^{2}+16\)

We can solve the system by solving the first equation for y and substituting the result in the second equation. x turns out to be equal to 4.2 and the area of a blue triangle would then be \(\frac{1}{2}(4)(4.2)=8.4\).

To find the area of a red triangle, we note that the rectangle has area 40 and consists of two blue and two red triangles. So the area of a red triangle is\(\frac{1}{2}(40-2(8.4))=11.6\).

Lastly, the pentagon would have area \(40-11.6=28.4\), since it lacks a red triangle as compared with the rectangle.

Have you tried cutting a rectangle to that size and playing with it?

I did that and I did get an answer. 

I doubt I did it the best way and my answer could be wrong because my working is on ridiculously scrappy bits of paper.

Thinking about it a little differently.

The area will be the same as the original rectangle minus triangle EFC

If you let FB=y

then you can work out that y=4.2  and  FC=5.8

You can work out that the triangle on the top is congruent to the triangle on the bottom 

the are of each of thes is 0.5*4*4.2 = 8.4

Add them together and you get  16.8u^2

The area of the original rectangle is 40u^2

If the area of the middle triangle is  T

then

40=16.8+2T 

T= 11.6u^2

so the area of the pentagon is     16.8+11.6 = 28.4 units squared

or

40-11.6 = 28.4 units square

[BCEF] = 1/2[ABCD]

ΔCDE ≅ ΔBCF

∠ECF = 2*arctan(4/10)

∠BCF = 90 - ∠ECF

BF = 4 * tan(∠ECF)

[BCDEF] = [BCEF] + 1/2(BC * BF)