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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