Two areas of similar triangles are 75 dm^2 and 48 dm^2 and the sum of their perimetres is 288 dm what is each triangles perimeter?

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Two areas of similar triangles are 75 dm^2 and 48 dm^2 and the sum of their perimetres is 288 dm what is each triangles perimeter?

2 thoughts on “Two areas of similar triangles are 75 dm^2 and 48 dm^2 and the sum of their perimetres is 288 dm what is each triangles perimeter?”

  1. Each triangle Perimeter is 128 dm and 160 dm.
    With similar geometric figures all around the dimensions, 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, 75 dm² and 48 dm²
    the a²:b² ratio is 25:16. Therefore, the ratio of the perimeters (1-D) would be √25: √16 or 5: 4.
    Solving, 288  = x + (5/4)x
    x = 128 dm
    Therefore the other triangle = 288 – 128 = 160 dm.
    One triangle Perimeter is 128 dm and the other is 160 dm.

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