The area of the parallelogram ABCD is 68. Point F is the midpoint of side AD. Find the area of the trapezoid AFCB.

Let the height of the parallelogram ABCD be equal to h, let us designate the side of the parallelogram BC as 2x.

The area of the parallelogram ABCD is equal to the product of the height and the base:

S (ABCD) = BC * h = 2xh = 68.

Hence xh = 68: 2; xh = 34.

Since F is the middle of side AD, then AF = AD = 1/2 * AD = x.

Let us express the area of the trapezoid AFCB (the area is equal to the product of the half-sum of the bases and the height):

S (AFCB) = (AF + BC) / 2 * h = (2x + x) / 2 * h = 3x / 2 * h = 3/2 * xh.

Hence S (AFCB) = 3/2 * 34 = 3 * 17 = 51.

Answer: The area of the trapezoid AFCB is 51.