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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Question : If $\triangle A B C \sim \triangle F D E$ such that $A B=9 \mathrm{~cm}, A C=11 \mathrm{~cm}, D F=16 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$, then the length of $BC$ is:
Question : $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$ and the perimeters of these triangles are 32 cm and 12 cm, respectively. If $\mathrm{DE}=6 \mathrm{~cm}$, then what will be the length of AB?
Question : In $\triangle \mathrm{ABC}$, $AB=20$ cm, $BC=7$ cm and $CA=15$ cm. Side $BC$ is produced to $D$ such that $\triangle \mathrm{DAB} \sim \triangle \mathrm{DCA}$. $DC$ is equal to:
Question : In a $\triangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $\triangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:
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