Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
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Let the two triangles be ABC and PQR.
We have:
$\triangle\text{ABC}\sim\triangle\text{PQR}$
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}$
$\triangle\text{ABC}\sim\triangle\text{PQR};$ therefore, their corresponding sides will be proportional.
$\Rightarrow\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\text{k }(\text{say})\dots(\text{i})$
$\Rightarrow\text{a}=\text{kp},\text{b}=\text{kq}$ and $\text{c}=\text{kr}$
$\therefore\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{kp+kq+kr}}{\text{p+q+r}}=\text{k}\dots(\text{ii})$
From (i) and (ii), we get:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}$
This completes the proof.
We have:
$\triangle\text{ABC}\sim\triangle\text{PQR}$
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}$
$\triangle\text{ABC}\sim\triangle\text{PQR};$ therefore, their corresponding sides will be proportional.
$\Rightarrow\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\text{k }(\text{say})\dots(\text{i})$
$\Rightarrow\text{a}=\text{kp},\text{b}=\text{kq}$ and $\text{c}=\text{kr}$
$\therefore\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{kp+kq+kr}}{\text{p+q+r}}=\text{k}\dots(\text{ii})$
From (i) and (ii), we get:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}$
This completes the proof.
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