# In parallelogram ABCM, the diagonals intersect at point O. The perimeter of triangle AOB is 17 cm

June 27, 2021 | education

| **In parallelogram ABCM, the diagonals intersect at point O. The perimeter of triangle AOB is 17 cm, BC = 9 cm, CM = 6 cm. Find the perimeter of triangle AOM.**

The opposite sides of the parallelogram are equal, which means AB = CM = 6 cm, AM = BC = 9 cm.

Consider the triangle AOB, express its perimeter:

P (AOB) = AB + BO + AO = 6 + AO + BO = 17.

Hence AO + BО = 17 – 6 = 11 (cm).

Let us express the perimeter of the triangle AOM:

P (AOM) = AM + AO + OM = 9 + (AO + OM).

Since the diagonals of the parallelogram intersect in the middle, BO = MO and AO + BO = AO + OM = 11 cm.

Therefore, the perimeter of the AOM triangle is:

P (AOM) = 9 + 11 = 20 (cm).

Answer: The perimeter of the AOM triangle is 20 cm.

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