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

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.