The continuation of the lateral sides AB and CD of the trapezoid ABCD intersect in the exact M.

| April 26, 2021 | education

The continuation of the lateral sides AB and CD of the trapezoid ABCD intersect in the exact M. Know the area of the trapezoid if BC: AD = 2: 5 and the area of the BMC is 12 cm2

Let us prove that triangle АМD is similar to triangle BMC.

The angle M in triangles is common, the angle MAD = MBC as the corresponding angles at the intersection of parallel straight lines BC and AD secant AM.

Then the triangles AMD and BMC are similar in two angles.

Let the length of the BC = 2 * X cm, then AD = 5 * X cm.

Let’s determine the coefficient of similarity of triangles. K = BC / AD = 2 * X / 5 * X = 2/5.

The areas of such triangles are referred to as the squared coefficient of their similarity.

Svms / Samd = 4/25.

Samd = 25 * Svms / 4 = 25 * 12/4 = 75 cm2.

Then Strap = Samd – Svms = 75 – 12 = 63 cm2.

Answer: The area of the trapezoid is 63 cm2.



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