Question

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

Answer 1

Let the two triangles be ABC and PQR.
We have:
$△ABC~△PQR$,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:
$\frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}$

$△ABC~△PQR$; therefore, their corresponding sides will be proportional.
$$\begin{gathered} \Rightarrow \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=k\left(\mathrm{say}\right)...\left(\mathrm{i}\right) \\ \Rightarrow a=kp,b=kq\mathrm{and}c=kr \\ \therefore \frac{\mathrm{Perimeter}\mathrm{of}△ABC}{\mathrm{Perimeter}\mathrm{of}△PQR}=\frac{a+b+c}{p+q+r}=\frac{kp+kq+kr}{p+q+r}=k...\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}=\frac{\mathrm{Perimeter}\mathrm{of}△ABC}{\mathrm{Perimeter}\mathrm{of}△PQR} \end{gathered}$$

This completes the proof.