Question : If $\triangle A B C \sim \triangle E D F$ such that $AB=6$ cm, $DF=16$ cm, and $DE=8$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 10 cm
Option 3: 14 cm
Option 4: 8 cm
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 12 cm
Solution : Given, $\triangle A B C \sim \triangle E D F$ such that $AB=6$ cm, $DF=16$ cm and $DE=8$ cm As $\triangle ABC$ ~ $\triangle EDF$ ⇒ $\frac{AB}{DE} = \frac{BC}{DF}$ ⇒ $\frac{6}{8} = \frac{BC}{16}$ $\therefore BC= 12$ cm Hence, the correct answer is 12 cm.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
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 : If in $\triangle$ABC, DE || BC, AB = 7.5 cm, BD = 6 cm, and DE = 2 cm then the length of BC in cm is:
Question : If $\triangle$ABC ~ $\triangle$DEF such that 2AB = DE and BC = 8 cm, then the length of EF is:
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:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile