What is the ratio of the perimiter of two equilateral tringles if their areas are 27 and 75?

Answer 1

We know that the perimeters of similar polygons have the same ratio as any two corresponding sides. Since we are dealing with equilateral triangles, we only need to find the scale factor because we know that in general, the ratio of the area of two similar figures is the square of the scale factor.

Let A1 = 75 and A2 = 27. So we have,

A1/A2 = 75/27 = 25/9

Thus, the scale factor or the ratio of any two sides or the ratio of perimeters of the two triangles is √25/√9 = 5/3.

Or find the side length of each equilateral triangle.

In a triangle A = bh/2.

Since we are dealing with equilateral triangles, we denote the bases of the triangles with 2x1 and 2x2, and the heights with x1√3 and x2√3. So that P1/P2 = 2x1/2x2 = x1/x2.

Let's find these side lengths.

A1 = b1h1/2 (replace A1 with 75, b1 = 2x1 and h1 =x1√3)

75 = (2x1)( x1√3)/2

75 = x12√3 (divide both sides by √3)

75/√3 = x12

A2 = b2h2/2 (replace A2 with 27, b2 = 2x2 and h2 =x2√3)

27 = (2x2)( x2√3)/2

27 = x22√3 (divide both sides by √3)

27√3 = x22

Since P1/P2 = x2/x2 also P12/P22 = x12/x22

So that,

P12/P22 = x12/x22

P12/P22 = (75/√3)/(27√3)

P12/P22 = 75/27 (square root of both sides)

P1/P2 = √75/√27 = √(75/27) = √(25/9) = 5/3