MCQ
$\triangle \text{ABC}$ is such that $\text{AB} =3 \ cm, \text{BC} =2 \ cm$ and $\text{CA}=2.5 \ cm$. If $\triangle \text{DEF} \sim \triangle \text{ABC}$ and $\text{EF} =4 \ cm$, then perimeter of $\triangle \text{DEF}$ is
  • A
    $30 \ cm$
  • B
    $7.5 \ cm$
  • C
    $22.5 \ cm$
  • $15 \ cm$

Answer

Correct option: D.
$15 \ cm$
Image
Since, $\triangle \text{DEF} \sim \triangle \text{ABC} [$Given$]$
$\Rightarrow \frac{A B}{D E}=\frac{B C}{E F}$
$\Rightarrow \frac{3}{D E}=\frac{2}{4}$
$\Rightarrow DE =6 \ cm$
Also, $\frac{A C}{D F}=\frac{B C}{E F}$
$\Rightarrow \frac{2.5}{D F}=\frac{2}{4}$
$\Rightarrow DF =5 \ cm$
$\therefore$ Perimeter of $\triangle \text{DEF=DE+EF+FD}$
$=6 \ cm+4 \ cm+5 \ cm$
$=15 \ cm$

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