The areas of two similar triangles are 72dm^2 and 50dm^2. The sum of their perimeters is 226dm. What is the perimeter of each of these triangles?

Answer 1

With similar geometric figures, consider the ratios of the dimensions:

length, 1-dimension: a:b

area, 2-dimensions: a²:b²

volume, 3-dimensions: a³:b³

Given the areas (2-D) of the two figures, 72 dm² and 50 dm², the a²:b² ratio is 36:25. Therefore, the ratio of the perimeters (1-D) would be √36:√25 or 6:5.

Therefore, the perimeter of the larger triangle is 6/5 the perimeter of the smaller triangle. Given the sum of the perimeters is 226 dm we can solve for the perimeters of both triangles.

Solving, 226 = x + (6/5)x

x = 102 8/11 or 102.73 dm

Smaller triangle perimeter, 102.73 dm; larger triangle perimeter, 123.27 dm