MCQ
$\triangle\text{ABC}$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $\triangle\text{DEF}\sim\triangle\text{ABC}$ and EF = 4cm, then perimeter of $\triangle\text{DEF}$ is:
- A7.5cm.
- ✓15cm.
- C22.5cm.
- D30cm.
✓
Answer
Correct option: B.
15cm.
$\triangle\text{DEF}\sim\triangle\text{ABC}$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$\triangle\text{s}$ 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 = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$\triangle\text{s}$ 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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