Let $$\triangle ABC \sim \triangle QPR$$ and $$\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{9}{16}$$. If AB = 12 cm, BC = 6 cm and AC = 9 cm, then QP is equal to:

As per the question,

$$\triangle ABC \sim \triangle PQR$$

$$\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{9}{16}$$

$$AB = 12$$

$$BC = 6$$

$$AC = 9 cm$$

We know that, if two triangles are similar then,

$$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{AB^2}{PQ^2}=\dfrac{BC^2}{QR^2}=\dfrac{AC^2}{PR^2}$$

Now substituting the values,

$$\Rightarrow\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{12^2}{PQ^2}=\frac{9}{16}$$

$$\Rightarrow \dfrac{144}{PQ^2}=\frac{9}{16}$$

Taking square root of both side,

$$\Rightarrow \dfrac{12}{PQ}=\dfrac{3}{4}$$

$$\Rightarrow PQ=\dfrac{12\times4}{3}$$

$$\Rightarrow PQ=16cm$$