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

The areas of two similar triangles are 50 dm sq and 32 dm sq, the sum of their perimeters is 117 cm What is the perimeter of the larger triangle?

First, let’s find the coefficient of similarity k of these two triangles.

In the initial data for this task, it is reported that the area of ​​the first triangle is 50 dm ^ 2, and the area of ​​the second is 32 dm ^ 2, therefore, the coefficient of similarity of these two triangles is:

k = √ (50/32) = √ (25/16) = √25 / √16 = 5/4.

Let us denote by p the perimeter of the smaller triangle.

Then the perimeter of the larger triangle should be 5p / 4, and since the sum of these perimeters is 117 dm, we can draw up the following equation:

p + 5p / 4 = 117,

solving which, we get:

9p / 4 = 117;

p = 117 * 4/9 = 4 * 117/9 = 4 * 13 = 52 dm.

Find the perimeter of the larger triangle:

5p / 4 = 5 * 52/4 = 5 * 13 = 65 dm.

Answer: 65 dm.