Line EF is parallel to the side AC of triangle ABC and cuts off a triangle from it, the sides of which

Line EF is parallel to the side AC of triangle ABC and cuts off a triangle from it, the sides of which are 6 times less than the sides of triangle ABC. Find the areas of these triangles if the area of the trapezoid is 70m2.

Since the segment EF is parallel to the AC side, the triangles ABC and EBF are similar in two angles. Angle B in triangles is common, angle BAC = BEF as the corresponding angles at the intersection of parallel straight lines AC and EF secant AB.

By condition, BE / AB = 1/6.

Then the coefficient of similarity of triangles is K = 1/6.

Savs / Svef = K2 = 1/36.

Savs = 36 * Svef. (one)

By condition, Saefс = 70 cm2.

Then Sаеf – Sеf = Sаеfc = 70 cm2. (2)

Let’s solve the system of equations 1 and 2.

36 * Svef – Svef = 70.

35 * Svef = 70.

Swef = 70/35 = 2 cm2.

Savs = 36 * Svef = 36 * 2 = 72 cm2.

Answer: The areas of the triangles are 2 cm2 and 72 cm2.