The shadow of the picture is a little misleading, but it is actually an area of' efhgd' p, a heptagon. This is the eye to do the problem. Don't be misled.

Secondly, triangles A'EP, D'PG and PFH are similar triangles. The way to prove it is to use parallel edges.

Let the rectangular height position a be AB = CD = D 'P = A 'P = A

It is known that the angle FPH is a right angle, and the triangle FPH is a triangle with the ratio of three sides of 3, 4 and 5. Specifically, PH=6, PF=8, FH= 10.

And BC=PF+FH+PH=8+ 10+6=24.

On triangle a' EP: a' p = a, then A'E=4/3a, EP=5/3a.

On the triangle d' pg: d "p = a, then D'G=3/4a, PG=5/4a.

ad = BC = AE+eg+GD = A ' e+eg+GD ' = A ' e+EP+PG+GD ' = 24。

Then 4/3a+5/3a+5/4a+3/4a=24.

That is, a=24/5.

I didn't understand it later. The shaded area is:

S triangle A'EP+S triangle D'PG+S trapezoid EGFH

Can you find the area of a right triangle?

The height of the trapezoid is a, and the upper and lower sides are also known. Simply calculate, it is 8 1.6.

If you don't understand, you can ask additional questions.

But because it is typing, there are really no mathematical symbols, so I can only barely understand them.