MCQ
$\triangle\text{ABC}\sim\triangle\text{DEF}.$ If $BC = 3\ cm, EF = 4\ cm$ and $\text{ar}(\triangle\text{ABC})=54\text{ cm}^2,$ then $\text{ar}(\triangle\text{DEF}):$
- A$108\ cm^2$
- ✓$96\ cm^2$
- C$48\ cm^2$
- D$100\ cm^2$
✓
Answer
Correct option: B.
$96\ cm^2$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\text{BC}=3\text{ cm},\ \text{EF}=4\text{ cm}$
$\text{ar}(\triangle\text{ABC})=54\text{ cm}^2$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{54}{\text{ar}(\triangle\text{DEF})}=\frac{3^2}{4^2}=\frac{9}{16}$
$\therefore\text{ar}(\triangle\text{DEF})=\frac{16\times54}{9}=96\text{ cm}^2$
$\text{BC}=3\text{ cm},\ \text{EF}=4\text{ cm}$
$\text{ar}(\triangle\text{ABC})=54\text{ cm}^2$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{54}{\text{ar}(\triangle\text{DEF})}=\frac{3^2}{4^2}=\frac{9}{16}$
$\therefore\text{ar}(\triangle\text{DEF})=\frac{16\times54}{9}=96\text{ cm}^2$
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