Question

In \( \triangle ABC \), \( DE \parallel BC \) such that \( \frac{AD}{DB} = \frac{3}{5} \). If \( AC = 5.6 \) cm, then \( AE \) is:

• A. \( 4.2 \) cm
• B. \( 3.1 \) cm
• C. \( 2.8 \) cm
• D. \( 2.1 \) cm

Hint:

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.

Answer 1

The Correct Option is C

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: \[ 5x = 3(5.6 - x) \] \[ 5x = 16.8 - 3x \] \[ 8x = 16.8 \] \[ x = 2.8 \]