The diagonals of a trapezoid are 15 and 20 cm. Find the area of a trapezoid if its middle line is 12.5 cm?

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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