The base AD of trapezoid ABCD is twice the base of the BC. A line m parallel to AB is drawn through the vertex C.

| February 9, 2021 | education

The base AD of trapezoid ABCD is twice the base of the BC. A line m parallel to AB is drawn through the vertex C. This line cuts off the CMD triangle from the trapezoid, the area of which is 3 square meters. see Find the area of the trapezoid.

Let the length of the base BC = X cm, then AD = 2 * X cm.

We will construct the height of the CH trapezium AВСD, which is also the height of the triangle СMD.

Since, by condition, CM is parallel to AB, then the quadrangle ABCM is a parallelogram, which means AM = BC = X cm, then DM = AD – AM = 2 * X – X = X cm.

The area of the triangle СMD is equal to: Ssmd = DM * CH / 2 = X * CH / 2 = 3 m2.

The area of the trapezoid is equal to: Savsd = (BC + AD) * CH / 2 = 3 * X * CH / 2.

Then Savsd = 3 * Ssmd = 3 * 3 = 9 cm2.

Answer: The area of the trapezoid is 9 cm2.



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