MCQ
In $\triangle\text{ABC},\text{DE }||\text{ BC}$ such that $\frac{\text{AD}}{\text{DB}}=\frac{3}{5}.$ AC = 5.6cm then AE =?
- A4.2cm
- B3.1cm
- C2.8cm
- ✓2.1cm
✓
Answer
Correct option: D.
2.1cm
In $\triangle\text{ABC},\text{DE }||\text{ BC}$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{AC}-\text{AE}}$
$\Rightarrow\frac{3}{5}=\frac{\text{AE}}{\text{5.6}-\text{AE}}$
$\Rightarrow3(5.6-\text{AE})=5\text{AE}$
$\Rightarrow16.8-3\text{AE})=5\text{AE}$
$\Rightarrow8\text{AE}=16.8$
$\Rightarrow\text{AE}=2.1\text{cm}$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{AC}-\text{AE}}$
$\Rightarrow\frac{3}{5}=\frac{\text{AE}}{\text{5.6}-\text{AE}}$
$\Rightarrow3(5.6-\text{AE})=5\text{AE}$
$\Rightarrow16.8-3\text{AE})=5\text{AE}$
$\Rightarrow8\text{AE}=16.8$
$\Rightarrow\text{AE}=2.1\text{cm}$
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