The sides of one triangle are equal to 9, 11, 12 cm. Its perimeter is 16 cm larger than the perimeter

The sides of one triangle are equal to 9, 11, 12 cm. Its perimeter is 16 cm larger than the perimeter of a similar triangle. Find the sides of the second triangle.

Consider a triangle ABC with sides AB = 9, BC = 11, AC = 12 cm.

Let triangle А1В1С1 be similar to triangle ABC with similarity coefficient K.

Then we can write the following equalities:

AB = K * A1B1, BC = K * B1C1, AC = K * A1C1.

Let’s denote the perimeter of the triangle ABC through P, and the perimeter of the triangle A1B1C1 through P1.

Therefore, we have:

P = AB + BC + AC = 9 + 11 + 12 = 32,

P1 = A1B1 + B1C1 + A1C1 = 1 / K * (AB + BC + AC) = K * P.

By the condition of the problem, it is known that

P – P1 = 16,

P – 1 / K * P = 16,

P * (1 – 1 / K) = 16,

32 * (1 – 1 / K) = 16,

1 – 1 / K = 1/2,

K = 2.

Hence,

9 = 2 * A1B1, 11 = 2 * B1C1, 12 = 2 * A1C1.

A1B1 = 4.5; B1C1 = 5.5; A1C1 = 6;