Question
In the given figure, DE || BC. Determine AC and AE.
✓
Answer
In the figure ABC is a triangle in which DE || BC
$\therefore\triangle\text{AED}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AE}}{\text{AB}}=\frac{\text{AD}}{\text{AC}}=\frac{\text{ED}}{\text{BC}}$
$\Rightarrow\frac{\text{AE}}{4}=\frac{14}{\text{AC}}=\frac{12}{15}$
Now, $\frac{\text{AE}}{4}=\frac{12}{15}\Rightarrow\text{AE}=\frac{12\times4}{15}$
$=\frac{4\times4}{5}=\frac{16}{5}$
and $\frac{14}{\text{AC}}=\frac{12}{15}\Rightarrow\text{AC}=\frac{14\times15}{12}=\frac{35}{2}$
Hence $\text{AC}=\frac{35}{2},\text{AE}=\frac{16}{5}$
$\therefore\triangle\text{AED}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AE}}{\text{AB}}=\frac{\text{AD}}{\text{AC}}=\frac{\text{ED}}{\text{BC}}$
$\Rightarrow\frac{\text{AE}}{4}=\frac{14}{\text{AC}}=\frac{12}{15}$
Now, $\frac{\text{AE}}{4}=\frac{12}{15}\Rightarrow\text{AE}=\frac{12\times4}{15}$
$=\frac{4\times4}{5}=\frac{16}{5}$
and $\frac{14}{\text{AC}}=\frac{12}{15}\Rightarrow\text{AC}=\frac{14\times15}{12}=\frac{35}{2}$
Hence $\text{AC}=\frac{35}{2},\text{AE}=\frac{16}{5}$
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