Given triangle ABC. Construct triangle A1B1C1, similar to triangle ABC, whose area is three times

Given triangle ABC. Construct triangle A1B1C1, similar to triangle ABC, whose area is three times the area of triangle ABC.

Let the area of the triangle ABC be equal to X cm2, Sаvs = X cm2, then, by condition, the area of the triangle А1В1С1 will be equal to: Sa1В1с1 = 3 * X cm2.

The ratio of the areas of similar triangles is equal to the square of the coefficient of similarity of the triangles.

Sa1v1s1 / Saavs = K2 = 3 * X / X.

K2 = 3.

K = √3.

Then A1B1 / AB = B1C1 / BC = A1C1 / AC = √3.

A1B1 = AB * √3 cm.

A1C1 = AC * √3 cm.

В1С1 = ВС * √3 cm.

Answer: The lengths of the sides of the triangle ABC need to be increased by √3 times.