The bisector of angle A of parallelogram ABCD intersects side BC at point K so that BK = 4

univerkov | June 9, 2021 | education

The bisector of angle A of parallelogram ABCD intersects side BC at point K so that BK = 4, CK = 2. find the area of the parallelogram if the angle A is 30 degrees.

Since AK is the bisector of angle A, then the angle BAK = DAK.

The BC side is parallel to AD, then the secant AR forms equal, criss-crossing angles KAD and BCA.

Then the angle BAK = BKA = DAK, and the triangle ABK is isosceles, AB = BK = 4 cm.

BC side length = (BK + CK) = 4 + 2 = 6 cm.

Then AD = BC = 6 cm.

Let’s calculate the area of the parallelogram. Savsd = AB * AD * SinBAD = 4 * 6 * Sin30 = 24 * 1/2 = 12 cm2.

Answer: The area of the parallelogram is 12 cm2.

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