Question : If $ riangle ABC \sim riangle QRP, \frac{\operatorname{area}( riangle A B C)}{\operatorname{area}( riangle 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

Question

Question : If $ riangle ABC \sim riangle QRP, \frac{\operatorname{area}( riangle A B C)}{\operatorname{area}( riangle 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

Team Careers360 · Accepted Answer

Correct Answer: 10 cm

Solution :
Given that $\frac{ ext{area}( riangle ABC)}{ ext{area}( riangle PQR)}=\frac{9}{4}$, BC = 15 cm, AB = 18 cm
$ riangle$ ABC ~ $ riangle$ QRP, hence the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
$\frac{ ext{area}( riangle ABC)}{ ext{area}( riangle PQR)}=\frac{BC^{2}}{PR^{2}}$
$⇒\frac{9}{4} = \frac{15^{2}}{PR^{2}}$
$⇒PR^{2}= \frac{15^{2} imes 4}{9}$
$⇒PR^{2}= \frac{900}{9}$
$⇒PR^{2}= 100$
$⇒PR= 10\ ext{cm}$
Hence, the correct answer is 10 cm.

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Question : If △ABC∼△QRP,area(△ABC)area(△QRP)=94,AB=18 cm,BC=15 cm, then the length of PR is:Option 1: 16 cmOption 2: 14 cmOption 3: 10 cmOption 4: 12 cm

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Question : If $△ A B C \sim △ Q R P, \frac{area (△ A B C)}{area (△ Q R P)} = \frac{9}{4}, A B = 18 cm, BC = 15 cm$ , then the length of $PR$ is:
Option 1: 16 cm
Option 2: 14 cm
Option 3: 10 cm
Option 4: 12 cm