In parallelogram ABCD, the bisectors of angles A and B meet at point K, which lies on side BC.

| May 14, 2021 | education

In parallelogram ABCD, the bisectors of angles A and B meet at point K, which lies on side BC. Find the perimeter of the parallelogram ABCD if AK = 9, DK = 12.

Since the sum of the angles BAD and ADC of the parallelogram is 180, and AK and DK are the bisectors of these angles, the sum of the angles DAK + ADK = 90, and then the triangle AKD is rectangular with a right angle AKD.

Then AD ^ 2 = KD ^ 2 + AK ^ 2 = 81 + 144 = 225.

DK = 15 cm.

Since AK and DK are the bisectors of the angles, the triangles ABK and CDK are isosceles. AB = VK, CD = SK, and since AB = CD are the opposite sides of the parallelogram, then AB = ВK = СK = CD.

Then BC = ВK + СK = AB + CD = 2 * AB = 2 * CD.

2 * AB = 15.

AB = 15/2 = 7.5 cm.

Determine the perimeter of the parallelogram.

P = 2 * (AB + AD) = 2 * (7.5 + 15) = 45 cm.

Answer: The perimeter of the parallelogram is 45 cm.



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