1)  We have right angled triangle ABE.  The hypotenuse AB is 61 units long; the leg AE is 60 units long.

    Hence, the leg BE is   = 11.



2)  We have right angled triangle ADE.  The hypotenuse AD is 156 units long; the leg AE is 60 units long.

    Hence, the leg ED is   = 144.



3)  The area of the triangle ABD is   =  =  = 4650 square units.



4)  Triangles ADE and BCD are similar (they are right-angled and have congruent angles DBC and BDA).

    From similarity, we have this proportion 

         = ,  or   =  = .

        hence  |DC| = .



5)  Then the area of the triangle BCD  is   =  =  = .




6)  The area of the trapezoid ABCD is the sum of areas of triangles  

         =  +  = 4650 +  =  = .




7)  From the similarity of triangles ADE and BCD we have this proportion

         = ,  or    = 

    
    From the proportion,  |BC| =  =  = 155.




8)  Now the area of the trapezoid ABCD is half sum of its bases AD and BC multiplied by the height of the trapezoid, or

         = .


    It gives  H =  = 59  = 59  = 59 .    ANSWER

AD ∥ CF  is given

AF ∥ CD  because they are both ⊥ DE

AFCD is a parallelogram because both pairs of opposite sides are ∥.

Use the Pythagorean theorem on right triangles △ADE and △ABE, and
we get DE = 144 an BE = 11.

△BEF ∽ △BDC  because  EF ∥ CD

 corresponding parts of similar triangles are in the
                  same ratio.





Solve that for BF and get



Then using the Pythagorean theorem in rt. △BEF, we get



Draw the altitude DG of of △BCD, DG ⊥ BC, where DG is 
the height of the trapezoid ABCD.

rt. △BDG ∽ rt. △BEF





Solve that for DG and get





Edwin