MCQ
In the given figure if $\angle\text{ADE}=\angle\text{ABC},\angle\text{ADE}=\angle\text{ABC},$ then $CE$ is equal to :
- A$3.$
- ✓$\frac{9}{2}$
- C$2.$
- D$5.$
✓
Answer
Correct option: B.
$\frac{9}{2}$
In $\triangle\text{ABC} $ and $\text{ ADE},$
$\triangle\text{ADE}=\angle\text{ABC}\ [$Given$]$
$\angle\text{A}=\angle\text{A} \ [$common$]$
$\therefore\triangle\text{ABC}\sim\triangle\text{ADE}\ [\text{AA}$ Similarity$]$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow \frac{2}{3}=\frac{3}{\text{EC}}$
$\Rightarrow \text{EC}=\frac{9}{2}\text{ cm}$
$\triangle\text{ADE}=\angle\text{ABC}\ [$Given$]$
$\angle\text{A}=\angle\text{A} \ [$common$]$
$\therefore\triangle\text{ABC}\sim\triangle\text{ADE}\ [\text{AA}$ Similarity$]$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow \frac{2}{3}=\frac{3}{\text{EC}}$
$\Rightarrow \text{EC}=\frac{9}{2}\text{ cm}$
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