Question

4. It is given that ∆ABC ~ ∆PQR with
BC/QR=1/3then
ar(∆PRQ)/
ar(∆BCA)=?​

Answer 1

Answer:

in traingle ABC and triangle PQR

AB=PQ

AC=PR

ANGLE PQR=ANGLE ABC

Answer 2

$\huge{\mathcal{\blue{\underline{Answer}}}}$

Theorem : the ratio of the areas of two similar triangle is equal to the square of the ratio of thier corresponding sides

$\frac{ar(triangle \: prq)}{ar(triangle \: bca)} = ( \frac{bc}{qr} ) {}^{2} \\ = (\frac{1}{3} ) {}^{2}$

$\huge{\mathcal{\blue{\underline{Area=1/9}}}}$