# In trapezoid ABCD, sides BC and AD are bases, AD = 2BC. Straight CM, parallel to AB, cuts off from the trapezoid

September 28, 2021 | education

| **In trapezoid ABCD, sides BC and AD are bases, AD = 2BC. Straight CM, parallel to AB, cuts off from the trapezoid triangle CMD, the area of which is 3 cm (2). Find the area of the trapezoid.**

Since CM, by condition, is parallel to AB, and BC is parallel to AM and AD as the base of a trapezoid, ABMD is a parallelogram, and therefore BC = AM.

Since AD = 2 * BC, then AM = DM = BC.

Let’s draw a diagonal AC.

In triangles, the AFM and DСМ bases are equal and the height of the CH is common, then Sasm = Ssdm = 3 cm2.

The AC diagonal divides the ABCM parallelogram into two equal triangles, then Savs = Sasm = 3 cm2.

Then Savsm = Savs + S asm + Sdsm = 3 + 3 + 3 = 9 cm2.

Answer: The area of the trapezoid is 9 cm2.

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