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

Question

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Answer 1

Let $X Y Z$ and $R S T$ be two similar triangle

then
$\frac{X Y}{R S} = \frac{X Z}{R T} = \frac{Y Z}{S T} - - - (1)$

The perimeter of $△ X Y Z = X Y + Y Z + X Z - - - (2)$

The perimeter of $△ R S T = R S + S T + R T - - - (3)$

from equation $(1)$

$\frac{X Y}{R S} = \frac{X Z}{R T} = \frac{Y Z}{S T} = \frac{X Y + X Z + Y Z}{R S + R T + S T}$

$= \frac{P e r i m e t e r o f △ X Y Z}{P e r i m e t e r o f △ R S T}$

from $(2)$ & $(3)$

$∴$ Hence proved.

Answer 2

Let

X Y Z

and

R S T

be two similar triangle

then

\frac{X Y}{R S} = \frac{X Z}{R T} = \frac{Y Z}{S T} - - - (1)

The perimeter of

△ X Y Z = X Y + Y Z + X Z - - - (2)

The perimeter of

△ R S T = R S + S T + R T - - - (3)

from equation

(1)

\frac{X Y}{R S} = \frac{X Z}{R T} = \frac{Y Z}{S T} = \frac{X Y + X Z + Y Z}{R S + R T + S T}

= \frac{P e r i m e t e r o f △ X Y Z}{P e r i m e t e r o f △ R S T}

from

(2)

&

(3)

∴

Hence proved.