The bisectors of angles A and B of the parallelogram ABCD intersect at point K lying on the side CD

univerkov | May 20, 2021 | education

The bisectors of angles A and B of the parallelogram ABCD intersect at point K lying on the side CD. Find the area of the parallelogram if BC = 12, and the distance from point K to the side AB = 4.

The bisectors ВK and СK cut off the isosceles triangles at the lateral sides of the parallelogram. In the triangle ВСК CB = СK, in the triangle АDК АD = AK, and since ВС = АD as opposite sides, then СK = DК = АD = ВС = 12 cm, then СD = DC + СK = 12 + 12 = 24 cm.

Determine the area of the triangle ВCК. Svsk = BC * KM / 2 = 12 * 4/2 = 12 cm2.

The area of the ВСK triangle is also equal to: Svsk = СK * ВН / 2, then 12 = 6 * ВН / 2.

BH = 24/6 = 4 cm.

Then Savsd = DC * ВН = 24 * 4 = 96 cm2.

Answer: The area of the parallelogram is 96 cm2.

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