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Area of a Triangle

Prove that th...

Question

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

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Solution

$L e t t h e t w o s i m i l a r △ l e s b e △ A B C & △ P Q R .$
$W h e n △ l e s a r e s i m i l a r,$
$1. T h e c o r r e s p o n d i n g a n g l e s a r e e q u a l .$
$2. T h e i r c o r r e s p o n d i n g s i d e s a r e p r o p o r t i o n a l .$
$h e n c e i n △ A B C & △ P Q R$
$\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$
$T h e p e r i m e t e r o f △ A B C = A B + B C + A C - (i)$
$T h e p e r i m e t e r o f △ P Q R = P Q + Q R + P R - (i i)$
$∴ \frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R} = \frac{A B + B C + A C}{P Q + Q R + P R}$
$= \frac{P e r i m e t e r o f △ A B C}{P e r i m e t e r o f △ P Q R}$

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Area of a Triangle

Standard IX Mathematics

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