Abcd is a trapezoid. what is the area of trapezoid?

Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
  Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
  Area = (1/2)(AB + AB/4)·h = (5/8)AB·h

The given dimensions let us determine the area of ∆BCE to be
  Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²

The total area of the trapezoid is also the sum of the areas ...
  Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
  Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
  Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h

Putting all of the above into the equation for the total area of the trapezoid, we have
  Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
  (5/8 -1/6 -1/12)AB·h = 30 cm²
  AB·h = (30 cm²)/(3/8) = 80 cm²

Then the area of the trapezoid is
  Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²