# In parallelogram ABCD, the diagonal is AC = 10. Find the area of the parallelogram ABCD

**In parallelogram ABCD, the diagonal is AC = 10. Find the area of the parallelogram ABCD if the angle BAC = 30 °, angle DAC = 45 °.**

Determine the values of the angles of the parallelogram.

Angle BAD = BAC + DAC = 30 + 45 = 750.

Angle ABC = 180 – BAD = 180 – 75 = 1050.

In triangle ABC we apply the theorem of sines.

ВС / Sin30 = AC / Sin105.

BC = AC * Sin30 / Sin105 = 10 * (1/2) / ((√3 + 1) / 2 * √2) = 10 * √2 / (√3 + 1) cm.

Angle BAC = CAD = 45, as cross-lying angles at the intersection of parallel straight lines BC and AD secant SA.

Determine the area of the triangle ABC.

Sас = ВС * АС * SinBCA / 2 = ((10 * √2 / (√3 + 1)) * 10 * √2 / 2) / 2 = 50 / (√3 + 1) cm2.

Since Savs = Sasd, then Savs = 2 8 Savs = 2 * 50 / (√3 + 1) = 100 / (√3 + 1) cm2.

Answer: The area of the parallelogram is 100 / (√3 + 1) cm2.