In ∆ABC, it is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm.


In ∆ABCit is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm. Also, in ∆DEF , EF = 8 cm and ∆DEF ∼ ∆ABC. Then, perimeter of ∆DEF is
(a) 22.5 cm
(b) 25 cm
(c) 27 cm
(d) 30 cm


(d) 30 cm

Perimeter of $\triangle A B C=A B+B C+C A=9+6+7.5=22.5 \mathrm{~cm}$

$\because \triangle D E F \sim \triangle A B C$

$\therefore \frac{\text { Perimeter } \triangle(A B C)}{\text { Perimeter }(\triangle D E F)}=\frac{B C}{E F}$

$\Rightarrow \frac{22.5}{\text { Perimeter }(\triangle D E F)}=\frac{6}{8}$

$\Rightarrow$ Perimeter $(\triangle D E F)=\frac{22.5 \times 8}{6}=30 \mathrm{~cm}$


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