$$\triangle$$ABC $$\sim$$ $$\triangle$$RQP and PQ = 10 cm, QR = 12 cm and RP = 16 cm. If ar($$\triangle$$PQR): ar ($$\triangle$$ABC) = $$\frac{9}{4}$$, then BC is equal to:

As per the question,

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

PQ = 10 cm, QR = 12 cm and RP = 16 cm

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

Now, by the similar triangle theorem,

ar($$\triangle$$RQP): ar ($$\triangle$$ABC) = $$(\frac{RQ}{AB})^2=(\frac{QP}{BC})^2=(\frac{PR}{CA})^2$$

Hence, substituting the values,

ar($$\triangle$$RQP): ar ($$\triangle$$ABC) = (\frac{QP}{BC})^2$$

$$\Rightarrow (\frac{10}{BC})^2=\dfrac{9}{4}$$

$$\Rightarrow \frac{100}{BC^2}=\dfrac{9}{4}$$

$$\Rightarrow BC=\sqrt{\dfrac{4}{9}\times100}=\dfrac{2\times 10}{3}=\dfrac{20}{3}$$cm