Question : If $\triangle ABC \sim \triangle QRP, \frac{\operatorname{area}(\triangle A B C)}{\operatorname{area}(\triangle Q R P)}=\frac{9}{4}, A B=18 \mathrm{~cm}, \mathrm{BC}=15 \mathrm{~cm}$, then the length of $\mathrm{PR}$ is:
Option 1: 16 cm
Option 2: 14 cm
Option 3: 10 cm
Option 4: 12 cm
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Correct Answer: 10 cm
Solution : Given that $\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)}=\frac{9}{4}$, BC = 15 cm, AB = 18 cm $\triangle$ ABC ~ $\triangle$ QRP, hence the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides. $\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)}=\frac{BC^{2}}{PR^{2}}$ $⇒\frac{9}{4} = \frac{15^{2}}{PR^{2}}$ $⇒PR^{2}= \frac{15^{2}\times 4}{9}$ $⇒PR^{2}= \frac{900}{9}$ $⇒PR^{2}= 100$ $⇒PR= 10\ \text{cm}$ Hence, the correct answer is 10 cm.
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