Find the area of a trapezoid given the length of a diagonal and the altitude of the trapezoid. In a trapezoid with perpendicular diagonals, the length of a diagonal is 5 and the length of the altitude of the trapezoid is 4. Find the area of the trapezoid.

I have tried using similar triangle properties and using pythagorean theorem but have found no solution. I tried using triangle BPC ~ triangle APD but didn't know where to go after that.
UPDATE: I have gotten to the point using the definition of the height of a triangle and the pythagorean theorem. But I don't know how to find the area of the top left triangle.enter image description here

I used the fact that ACF and ABC are congruent by ASA so I got the area of the trapezoid is 38 but I'm not sure if that is right
 A: 
Hint: Use this figure to find:
1-In a trapzoid if diagonals are perpendicular then one side is equal to one diagonal(AB=BD).
2- In this case AD=6 (use Pythagoras's theorem).
3-Use the congruence of right angled triangles with vertex E to find the measure of side BC.
4-The area will be $\frac{(BC+AD)}2\times BH\approx 17$.
A: 
Say diagonal $AC = 5$. As diagonals are perpendicular to each other, $\angle AOD = 90^\circ$
$\triangle AOD \sim \triangle AFC$ and $\triangle AOD \sim \triangle COB$. If $OC = y, AO = 5 - y$,
$ \displaystyle \frac{BC}{y} = \frac{AD}{5-y} = \frac{5}{3}$
$ \implies \displaystyle BC = \frac{5y}{3}, AD = \frac{25-5y}{3}$
$ \displaystyle BC + AD = \frac{25}{3}$
Area of trapezoid is,
$ \displaystyle A = \frac{1}{2} (BC + AD) \cdot CF = \frac{50}{3}$
A: 
Let $BF=4$ and $AC=5$
Draw $BE\perp BD$, so $BE \parallel AC$ and $BE=AC=5$
Area of trapezoid$=\frac{(AD+BC)\times BF}{2}=\frac{(EA+AD)\times BF}{2}=\frac{ED\times BF}{2}=\frac{\frac{25}{3}\times 4}{2}=\frac{50}{3}$, since $EA=BC$
Applying Pytha
$EF^2=BE^2-BF^2$, $EF=3$
Applying Euclidean Theorems
$BE^2=EF\times ED$ then $25=3\times ED=$, $ED=\frac{25}{3}$
