In the parallelogram ABCD, the angle ABC is 150. Find the area of this parallelogram if the bisector

In the parallelogram ABCD, the angle ABC is 150. Find the area of this parallelogram if the bisector of angle A intersects the side BC at point M and BM = 8 MC = 6.

1. Draw line MN parallel to line AB. The parallelogram ABMN is a rhombus, since the diagonal AM is the bisector of the parallelogram:

AB = BM = 8.

2. The sum of the one-sided angles is 180 °:

∠A + ∠B = 180 °;

∠A = 180 ° – ∠B = 180 ° – 150 ° = 30 °.

3. In a right-angled triangle ABH, the leg BH, which lies opposite an angle of 30 °, is equal to half of the hypotenuse AB:

BH = 1/2 * 8 = 4.

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

S = BH * BC = 4 * (8 + 6) = 4 * 14 = 56.

Answer: 56.