Question

Question in the attachment !!​

Answer 1

ar.∆ABC/ar.∆PQR=

$\frac{bc {}^{2} }{pq {}^{2} }$

=

$( \frac{bc}{pq} ) {}^{2}$

=

$( \frac{1}{3} ) {}^{2}$

=

$\frac{1}{9}$