The middle line of the MC of the triangle ABC cuts off the triangle MВK from it, the area of which is 10 cm2. Find the area of the triangle ABC.
Let us prove that triangles ABC and CME are similar.
The angle B in triangles is common, the angle BAC is equal to the ВKM triangle as the corresponding angles at the intersection of parallel straight lines KM and AC secant AB. Then the ABC triangle is similar to the ВKM triangle in two angles.
KM is the middle line of the triangle, which means its length is half the base of the AC.
Then the coefficient of similarity of triangles is: K = AC / KM = 2 * KM / KM = 2.
Since the ratio of the areas of similar triangles is equal to the square of the similarity coefficient, then:
Savs / Svkm = 2 ^ 2.
Savs = 4 * 10 = 40 cm2.
Answer: The area of triangle ABC is 40 cm2.
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