Draw a straight line SC parallel to BD through the vertex C of the trapezoid.
The formed quadrilateral BCKD is a parallelogram, since its opposite sides are parallel. Then DK = BC, CK = BK.
The segment AD = AD + DK = AD + BC, that is, it is equal to the sum of the lengths of the bases of the trapezoid.
Since MP = (BC + AD) / 2, then (BC + AD) = AK = 2 * MP = 2 * 12.5 = 25 cm.
Since the height of the trapezoid is equal to the height of the triangle ACK, the area of the triangle ACK is equal to the area of the trapezoid ABCD.
Let us determine the area of the triangle ACK by Heron’s theorem.
Sask = √p * (p – AC) * (p – CK) * (p – AK), where p is the semiperimeter of the triangle.
P = (AC + CK + AK) / 2 = (15 + 20 + 25) / 2 = 30 cm.
Sask = √30 * 15 * 10 * 5 = √22500 = 150 cm2.
Savsd = Sask = 150 cm2.
Answer: The area of the trapezoid is 150 cm2.
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