In parallelogram ABCD, the sides AB and AD are respectively equal to 13 and 20. Point M is marked on the side AD so that AM = MD, BA = BM. Find the area of triangle ABM if the area of parallelogram ABCD is known to be 240.
By condition, AM = MD, then AM = MD = AD / 2 = 20/2 = 10 cm.
Then, Ravs = 13 + 13 + 10 = 36, and p = 36/2 = 18 cm.
According to Heron’s theorem: Savm = √18 * (18 – 13) * (18 – 13) * (18 – 10) = √18 * 5 * 5 * 8 = √3600 = 60 cm2.
Answer: The area of the triangle ABM is 60 cm2.
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