In the parallelogram ABCD above, AD = 12. If the area of triangle ABE

Area of triangle ABE = 1/2*BE*AG
Area of parallelogram ABCD = AD* BF
Given 1/2*BE*BF=3/8*AD*BF
But AG=BF and AD=12
=> 1/2*BE*BF=3/8*AD*BF
=> BE=9
CE=12-9=3

We can let the height of the parallelogram be 10. Notice that this will the be height of the three triangles we see in the diagram.

Therefore, the parallelogram has an area of 12 x 10 = 120 and hence the area of triangle ABC = 3/8 x 120 = 45. Triangle AED is half of the area of the parallelogram; thus, it has an area 1/2 x 120 = 60. Therefore, the area of triangle EDC is 120 - 45 - 60 = 15, and we can create the following equation to find EC:

1/2 x EC x 10 = 15

5 x EC = 15

EC = 3
Alternate Solution:

Let the height of the parallelogram be h. Then, the area of the parallelogram is 12h.

Notice that the height of triangle ABE is also h and, in terms of |BE| and h, the area of the triangle ABE is (|BE|*h)/2. We know that this is equal to 3/8 of 12h, so let’s set up the following equation:

(|BE|*h)/2 = (3/8)*12h

|BE|*h = (3/4)*12h = 3 * 3h = 9h

|BE| = 9

Since |BE| = 9, |EC| = |BC| - |BE| = |AD| - |BE| = 12 - 9 = 3.

Answer: D

ABCD is para
ABE is triangle 3/8 para
AECD is a trapezium 5/8 para

Area para = bh = 12h
Area trapezium = b1+b2/2*h… 12+EC/2*h = 5/8*12h
…12+EC/2=5/8*12…12+EC=60/8*2…
EC=60/4-12=15-12=3

Ans (D)

Height of triangle = height of parallelogram = H