In parallelogram ABCD, diagonal BD is perpendicular to side AB, one of the parallelogram angles = 120

| September 7, 2021 | education

In parallelogram ABCD, diagonal BD is perpendicular to side AB, one of the parallelogram angles = 120 °, AD = 12cm, O is the point of intersection of the diagonals. Find the diagonals of a parallelogram and the area ΔCDO.

Since the sum of the adjacent angles of the parallelogram is 180, then the angle BAD = 180 – 120 = 60, then the angle BDA = 180 – 90 – 60 = 30. In a right-angled triangle ABD, the leg AB lies opposite the angle 30, then AB = AD / 2 = 12 / 2 = 6 cm.

In the triangle ABD, by the cosine theorem, we define the length of the segment BD.

BD ^ 2 = AB ^ 2 + AD ^ 2 – 2 * AB * AD * Cos60 = 36 + 144 – 2 * 6 * 12 * (1/2) = 180 – 72 = 108.

ВD = 6 * √3 cm.

Then, since the diagonals of the parallelogram are halved at the intersection point, then BO = DO = BD / 2 = 6 * √3 / 2 = 3 * √3 cm.

Determine the area of ​​the triangle AOB.

Saov = AB * ВO / 2 = 6 * 3 * √3 / 2 = 9 * √3 cm2.

Since the triangles AOB and COD are equal on three sides, then Scod = Saov = 9 * √3 cm2.

Answer: The area of ​​the CDO triangle is 9 * √3 cm2.



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