MCQ
$\triangle\text{ABC}$ is such that $AB = 3\ cm, BC = 2\ cm$ and $CA = 2.5\ cm$. If $\triangle\text{DEF}\sim\triangle\text{ABC}$ and $EF = 4\ cm,$ then perimeter of $\triangle\text{DEF}$ is :
- A$7.5\ cm.$
- ✓$15\ cm.$
- C$22.5\ cm.$
- D$30\ cm.$
✓
Answer
Correct option: B.
$15\ cm.$
$\triangle\text{DEF}\sim\triangle\text{ABC}$
$AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm, EF = 4\ cm.$
$\triangle\text{s}$ are similar
$\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}}$
$\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5}$
Now $\frac{\text{DE}}{3}=\frac{4}{2}$
$\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{ cm}$
and $\text{FD}=\frac{4}{2}$
$\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{ cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$=6+4+5=15\text{ cm}.$
$AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm, EF = 4\ cm.$
$\triangle\text{s}$ are similar
$\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}}$
$\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5}$
Now $\frac{\text{DE}}{3}=\frac{4}{2}$
$\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{ cm}$
and $\text{FD}=\frac{4}{2}$
$\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{ cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$=6+4+5=15\text{ cm}.$
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