Issues with a 3D Trigonometry Problem I have the following shape:

Where the midpoint of $EF$ lies vertically above the intersection of the diagonals $AC$ and $BD$.
Now, I need to find the heights of the trapezium and triangles in order to calculate the total surface area of the sloping faces.
This was my working, but I keep getting an incorrect answer:

*

*For the height of the triangle:

I did $tan(50)$ = $\frac{Height_{TRIANGLE}}{3.5}$
$Height_{TRIANGLE}$ = $3.5 * tan(50)$


*For the height of the trapezium:

The hypotenuse of the triangle = the length of the slanted sides of the trapezium
$cos(50)$ = $\frac{3.5}{Hypotenuse_{TRIANGLE}}$
$Hypotenuse_{TRIANGLE}$ = $\frac{3.5}{cos(50)}$
And now to find the height of the trapezium.
If I made a triangle with the height of the trapezium as the opposite side.
$sin(50)$ = $\frac{Height_{TRAPEZIUM}}{Hypotenuse_{TRIANGLE}}$
$Height_{TRAPEZIUM}$ = $sin(50)$ * $Hypotenuse_{TRIANGLE}$
$Height_{TRAPEZIUM}$ = $sin(50)$ * $\frac{3.5}{cos(50)}$ = 3.5 * $\frac{sin(50)}{cos(50)}$ = $3.5 * tan(50)$
What am I missing here?
 A: 
Here is your diagram with a few extra lines added in blue and a few extra points labeled. Now there are some right triangles in the figure and you can start applying trigonometric functions to them.
For example,
$$ \tan(50^\circ) = \frac{FH}{HK}. $$
Accurate, precise labeling helps avoid errors.
For example, what is "$Height_{TRIANGLE}$"? Some candidates are the length of segment $FG$ (the most likely candidate, as $\triangle CDF$ is one of the few triangles in the original figure and $FG$ is the altitude from the base $CD$)
and the length of segment $FH$ (which is literally the "height" of the farthest point of
$\triangle CDF$ above the plane $ABCD$, and which your formula implies is the intended length).
You can avoid this confusion by simply writing $FG$ when you mean the length of the segment $FG$ and $FH$ when you mean the length of the segment $FH.$
The other advantage to precise, accurate labeling is that other people have a chance to actually understand what you're trying to say.
A: The height of both trapezium and triangle is the same and you only need to calculate the hypotenuse of triangle $PQR$.
You have the base and the angle.
$h = \dfrac{3.5}{\cos(50^\circ)}≈5.445m$

A: Assuming $ABCD$ is a rectangle whose center $P$ is right below midpoint $O$ of $EF$ which edge is parallel to base plane $ABCD.$
I suppose you started with a square pyramid 7 m base $ abcdO $ and extended either side faces keeping symmetry with 1.5 meters parallel translated/shifted colored planes shown. The red edges are parallel in this translation/shift.
It  makes no difference to the central right triangle $OPQ$. Height $h$ is retained the same.
Accordingly from this right triangle $OPQ$ the height of the trapezium and two triangles is
$$ h=3.5 \tan 50^{\circ};$$

