MCQ
In the given figure if $\triangle\text{AED}\sim\triangle\text{ABC},$ then $DE$ is equal to :
- ✓$5.6\ cm.$
- B$6.5\ cm.$
- C$7.5\ cm.$
- D$5.5\ cm.$
✓
Answer
Correct option: A.
$5.6\ cm.$
Since $\triangle\text{AED}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AE}}{\text{AB}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{12}{16+14}=\frac{\text{DE}}{14}$
$\Rightarrow\text{DE}=\frac{12\times14}{30}=\frac{84}{15}=5.6\text{ cm}$
$\therefore\frac{\text{AE}}{\text{AB}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{12}{16+14}=\frac{\text{DE}}{14}$
$\Rightarrow\text{DE}=\frac{12\times14}{30}=\frac{84}{15}=5.6\text{ cm}$
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