Given two similar triangles one of which has twice the perimeter of the other, by what factor is the area of the larger triangle bigger than the smaller ?

2

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4

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\sqrt{2}

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2 \sqrt{2}

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Solution

The correct option is B $4$
$L e t △ A B C a n d △ P Q R b e s m a l l e r a n d b i g g e r t r i a n g l e s G i v e n △ A B C \sim △ P Q R 2 P e r i m e t e r o f △ A B C = P e r i m e t e r o f △ P Q R \Rightarrow \frac{P e r i m e t e r o f △ A B C}{P e r i m e t e r o f △ P Q R} = \frac{1}{2} T w o s i m i l a r t r i a n g l e s w h o h a v e s c a l e f a c t o r o f a : b t h e n r a t i o o f t h e r e a r e a s w i l l b e a^{2} : b^{2} \Rightarrow \frac{A r e a o f △ A B C}{A r e a o f △ P Q R} = ({\frac{1}{2})}^{2} = \frac{1}{4} \Rightarrow A r e a o f △ P Q R = 4 \times A r e a o f △ A B C ∴ A r e a o f l a r g e r t r i a n g l e i s 4 t i m e s b i g g e r t h a n a r e a o f s m a l l e r t r i a n g l e . H e n c e o p t i o n (B) i s c o r r e c t a n s w e r$

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