The bisector of angle B of parallelogram ABCD intersects side AD at point K so that AK = 6, AD = 8.

The bisector of angle B of parallelogram ABCD intersects side AD at point K so that AK = 6, AD = 8. Find the area of the parallelogram if the angle B is 150 degrees

Corners B and C – one-sided inner:

<B + <C = 180 °;

<C = 180 ° – <B = 180 ° – 150 ° = 30 °.

The opposite angles of the parallelogram are:

<A = <B = 30 °.

The bisector BG divides <B in half:

<ABG = <CBG = <B / 2 = 150 ° / 2 = 75 °;

<AGB = 180 ° – <A – <ABG = 180 ° – 30 ° – 75 ° = 75 °;

<ABG = <AGB.

It follows from the equality of the angles that ΔABG is isosceles:

AG = AB = 6 cm.

The height BH of the parallelogram is omitted from point B. ΔABH is rectangular. The BH leg is opposite the angle A = 30 °:

BH = AB / 2 = 6/2 = 3 cm.

SABCD = AD * BH = 8cm * 3cm = 24cm ^ 2.

Answer: 24 cm ^ 2.