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

May 28, 2021 | education

| **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.

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