MCQ
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $DE = 3\ cm, EF = 2\ cm, DF = 2.5\ cm, BC = 4\ cm,$ then perimeter of $\triangle\text{ABC}$ is :
- A$18\ cm.$
- B$20\ cm.$
- C$12\ cm.$
- ✓$15\ cm.$
✓
Answer
Correct option: D.
$15\ cm.$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$DE = 3\ cm, EF = 2\ cm, DF = 2.5\ cm, BC = 4\ cm$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$= DE + EF + DF$
$= 3 + 2 + 2.5 = 7.5\ cm$
Now $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}$
$=\frac{\text{AB}+\text{BC}+\text{CA}}{\text{DE}+\text{EF}+\text{DF}}$
$=\frac{4}{2}=\frac{\text{AB}+\text{BC}+\text{CA}}{7.5}$
$\Rightarrow\text{AB}+\text{BC}+\text{CA}=\frac{4\times7.5}{2}=15$
$\therefore$ Perimeter of $\triangle\text{ABC}=15\text{ cm}.$
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