Question

Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Answer 1

Given: Two triangles ΔABC and ΔPQR are similar. Let BC = a, CA = b, AB = c and QR = p, RP = q, PQ = r.

To Prove: The ratio of their perimeters equals the ratio of any pair of corresponding sides, i.e. `a/p = b/q = c/r = (a + b + c)/(p + q + r)`.

Proof [Step-wise]:

1. From ΔABC ∼ ΔPQR, corresponding sides are proportional.

∴ `a/p = b/q = c/r = k`   ...(For some constant k)

2. From (1) we have a = k·p, b = k·q, c = k·r.

3. Perimeter of ΔABC = a + b + c

= k·p + k·q + k·r

= k(p + q + r)

4. Therefore `(a + b + c)/(p + q + r) = k`.

5. Combining with (1) gives `a/p = b/q = c/r = (a + b + c)/(p + q + r)`.

Hence the ratio of the perimeters of two similar triangles equals the ratio of their corresponding sides.