In the given figure, $△ A B C \sim △$ PQR and AM and PN are medians of $△$ ABC and $△$ PQR respectively. Then $\frac{A r e a o f △ A B C}{A r e a o f △ P Q R}$ = _____

$\frac{{A B}^{2}}{{P Q}^{2}}$

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$\frac{{A M}^{2}}{{P N}^{2}}$

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(a) and (b) above.

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\frac{{P Q}^{2}}{{Q R}^{2}}

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Solution

The correct option is C (a) and (b) above.
We are given two triangles ABC and PQR such that $△$ ABC $\sim$ $△$ PQR.

So, $\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$

$\frac{A B}{P Q} = \frac{2 B M}{2 Q N} = \frac{B M}{Q N}$

$\Rightarrow \frac{A B}{P Q} = \frac{B M}{Q N}$

$∠$ B = $∠$ Q

$∴ △ A B M \sim △$ PQN (by SAS similarity criterion)

$\Rightarrow \frac{A B}{P Q} = \frac{A M}{P N}$

So , $\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{A B^{2}}{P Q^{2}} = \frac{A M^{2}}{P N^{2}}$

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