The bisector of one of the corners divides the parallelogram into two parts, the difference of the perimeters of which is 10 cm.

| September 6, 2021 | education

The bisector of one of the corners divides the parallelogram into two parts, the difference of the perimeters of which is 10 cm. Find the perimeter of the parallelogram if the sides of the parallelogram are 4: 9.

Let the length of the smaller side of the parallelogram be 4 * X cm, then the length of the larger side is 9 * X cm.

Since BM is the bisector of the angle ABC, the triangle ABM is isosceles, AM = AB = 4 * X cm.

Then DM = AD – AM = 9 * X – 4 * X = 5 * X.

The perimeter of the triangle ABM is equal to: Ravm = (AB + BM + AM) = (4 * X + BM + 4 * X) = (8 * X + BM).

The perimeter of the BCDM quadrangle is: Pvsdm = (BC + CD + DM + BM) = (9 * X + 4 * X + 5 * X + BM) = (18 * X + BM).

By condition, Rvsdm – Ravm = 10.

18 * X + BM – 8 * X – BM = 10 cm.

10 * X = 10.

X = 10/10 = 1 cm.

AB = CD = 1 * 4 = 4 cm.

BC = AD = 1 * 9 = 9 cm.

Then Ravsd = 2 * (AB + BC) = 2 * 13 = 26 cm.

Answer: The perimeter of the parallelogram is 26 cm.



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