$$\triangle$$ABC $$\sim$$ $$\triangle$$RQP and PQ = 10 cm, QR =12 cm and RP = 18 cm. If ar($$\triangle$$ABC) : ar($$\triangle$$RQP) = $$\frac{4}{9}$$, then AB is equal to:

Given the question,

$$\triangle$$ABC $$\sim$$ $$\triangle$$RQP

PQ = 10 cm, QR =12 cm and RP = 18 cm

ar($$\triangle$$ABC) : ar($$\triangle$$RQP) = $$\frac{4}{9}$$

As per the rule of similar triangle,

We know that ar($$\triangle$$ABC) : ar($$\triangle$$RQP)$$=(\dfrac{AB}{RQ})^2=(\dfrac{BC}{QP})^2=(\dfrac{CA}{RP})^2$$

Now, substituting the values

$$\Rightarrow \dfrac{ar(\triangle ABC)}{ar(\triangle RQP)}=(\dfrac{AB}{12})^2$$

$$\Rightarrow (\dfrac{AB}{12})^2=\dfrac{4}{9}$$

Taking square root of both side,

$$\Rightarrow \dfrac{AB}{12}=\dfrac{2}{3}$$

$$\Rightarrow AB=\dfrac{2}{3}\times12$$

$$\Rightarrow AB=8$$cm