Triangle $A B C$ has sides of length $5, 6 a n d 7 u n i t s$ while $△ P Q R$ has a perimeter of $360 u n i t s$ . If $△ A B C$ is similar to $△ P Q R$ , then find the sides of $△ P Q R$ .

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Solution

Given, AB= 5 , AC = 6 and BC = 7. Perimeter( $△ P Q R$ ) = 360
Now, Perimeter( $△ A B C$ ) = 5 + 6 + 7 = 18 units
hence, $\frac{P e r i m e t e r (△ A B C)}{P e r i m e t e r (△ P Q R)} = \frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$
$\frac{18}{360} = \frac{5}{P Q} = \frac{7}{Q R} = \frac{6}{P R}$
$\frac{1}{20} = \frac{5}{P Q} = \frac{7}{Q R} = \frac{6}{P R}$
$P Q = 100$ units, $Q R = 140$ units, $P R = 120$ units

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