If $$\triangle ABC \sim \triangle QPR, \frac{ar(ABC)}{ar(\triangle PQR)} = \frac{9}{4}, AC = 12 cm, AB = 18 cm$$ and $$BC == 15 cm$$ then $$PR$$ is equal to:
ABC is similar to PQRÂ
Now area of ABC : area of PQR = 9:4 = (side ratio )^2
so we get side ratio = 3:2
so AC:PR =3:2
so we get PR =8cm
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