In parallelogram ABCD, BC = 3√3, angle BAD = 30 degrees, BD = BC, Find the area of ABCD.

Since, by condition, the diagonal ВD is equal to the side ВС, the triangle ВСD is isosceles, in which the angle ВСD = ВDC = 30.

Let’s build the height BH of triangle BD, which is its median, then CH = DH = CD / 2, and triangles BCH and BDH are rectangular.

In a right-angled triangle ВСН, СН = ВС * CosВСН = 3 * √3 * Cos30 = 3 * √3 * √3 / 2 = 4.5 cm.

Then CD = 2 * CH = 2 * 4.5 = 9 cm.

The parallelogram area is equal to: Savsd = ВС * СD * Sin30 = 3 * √3 * 9 * 1/2 = 13.5 * √3 cm2.

Answer: The area of the parallelogram is 13.5 * √3 cm2.