# ABCD – trapezoid BC = 2 cm AD = 8 cm AC = 4cm – diagonal BC, AD base.

ABCD – trapezoid BC = 2 cm AD = 8 cm AC = 4cm – diagonal BC, AD base. In what ratio the AC divides the area of the trapezoid.

Consider triangles ABC and ACD formed by the diagonal AC.

The ACB angle is equal to the CAD angle as the intersecting angles at the intersection of parallel lines BC and AD secant AC.

Let the angle ACB and CAD be equal to α.

Then the area of the triangle ACD will be equal to Sacd = AC * AD * Sinα = 4 * 8 * Sinα = 32 * Sinα.

The area of the triangle ABC is equal to: Sавс = AC * BC * Sinα = 4 * 2 * Sinα = 8 * Sinα.

Then Saws / Sasd = 8 * Sinα / 32 * Sinα = 1/4.

Answer: The area ratio is 1/4.

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