Question:
In \( \triangle ABC \), \( DE \parallel BC \) such that \( \frac{AD}{DB} = \frac{3}{5} \). If \( AC = 5.6 \) cm, then \( AE \) is:
In \( \triangle ABC \), \( DE \parallel BC \) such that \( \frac{AD}{DB} = \frac{3}{5} \). If \( AC = 5.6 \) cm, then \( AE \) is:
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The **Basic Proportionality Theorem** states that if a line is parallel to one side of a triangle, it divides the other two sides in the same ratio.
Updated On: Oct 27, 2025
- \( 4.2 \) cm
- \( 3.1 \) cm
- \( 2.8 \) cm
- \( 2.1 \) cm
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The Correct Option is C
Solution and Explanation
Since \( DE \parallel BC \), by the Basic Proportionality Theorem (Thales' Theorem), we get:
\[ \frac{AD}{DB} = \frac{AE}{EC} \] Given \( \frac{AD}{DB} = \frac{3}{5} \), this means:
\[ \frac{AE}{EC} = \frac{3}{5} \] Since \( AC = AE + EC = 5.6 \), let \( AE = x \), then \( EC = 5.6 - x \).
\[ \frac{x}{5.6 - x} = \frac{3}{5} \] Cross multiplying: \[ 5x = 3(5.6 - x) \] \[ 5x = 16.8 - 3x \] \[ 8x = 16.8 \] \[ x = 2.8 \]
\[ \frac{AD}{DB} = \frac{AE}{EC} \] Given \( \frac{AD}{DB} = \frac{3}{5} \), this means:
\[ \frac{AE}{EC} = \frac{3}{5} \] Since \( AC = AE + EC = 5.6 \), let \( AE = x \), then \( EC = 5.6 - x \).
\[ \frac{x}{5.6 - x} = \frac{3}{5} \] Cross multiplying: \[ 5x = 3(5.6 - x) \] \[ 5x = 16.8 - 3x \] \[ 8x = 16.8 \] \[ x = 2.8 \]
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