Question : If in a $\triangle$ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $\frac{AD}{BD}$ = $\frac{3}{5}$. If AC = 4 cm, then AE is:
Option 1: 1.5 cm
Option 2: 2.0 cm
Option 3: 1.8 cm
Option 4: 2.4 cm
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Correct Answer: 1.5 cm
Solution :
Given: $\frac{AD}{BD}$ = $\frac{3}{5}$
In $\triangle$ABC and $\triangle$DBE,
$\angle$BAC = $\angle$DAE (same angle)
$\angle$ADE = $\angle$ABC (corresponding angles)
$\angle$AED = $\angle$ACB (corresponding angles)
By AAA similarity, $\triangle$ABC ~ $\triangle$ADE
⇒ $\frac{AD}{AB}$ = $\frac{AD}{AD+BD}$ = $\frac{3}{3+5}$ = $\frac{3}{8}$
Now, $\frac{AD}{AB}$ = $\frac{AE}{AC}$
⇒ $\frac{3}{8}$ = $\frac{AE}{4}$ [since AC = 4 cm]
⇒ AE = $\frac{3×4}{8}$ = 1.5 cm
Hence, the correct answer is 1.5 cm.
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