Let $$\triangle ABC \sim \triangle QPR and \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{9}{4}$$. If AB = 12 cm, BC = 6 cm and AC = 9 cm,then QR is equal to:Â
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
$$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AC^2}{QR^2}$$
$$\frac{9}{4}=\frac{81}{QR^2}$$
QR=6.
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