The areas of two similar triangles are 50dm² and 32dm², and the sum of their perimeters is 117 dm.

The areas of two similar triangles are 50dm² and 32dm², and the sum of their perimeters is 117 dm. Find the perimeter of each triangle.

The ratio of the areas of similar triangles is equal to the square of the similarity coefficient. Let’s find the coefficient of similarity of similar triangles:

k² = 50/32 = 1.5625;

k = √1.5625;

k = 1.25.

The ratio of the respective sides of such triangles, as well as the ratio of their perimeters, is equal to the coefficient of similarity.

Let’s solve the problem using the equation, where:

x is the perimeter of one triangle;

(117 – x) – the perimeter of the second triangle;

Let’s compose and solve an equation in which the ratio of the perimeters of similar triangles is equal to the coefficient of similarity:

(117 – x) / x = 1.25;

125 * x = 100 * (117 – x);

125x = 11700 – 100x;

225x = 11700;

x = 11,700 / 225;

x = 52 dm – perimeter of the first triangle;

117 – x = 117 – 52 = 65 dm – the perimeter of the second triangle.

Answer: 52 dm, 65 dm