Question:

In \(\triangle ABC, DE || BC\). If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1) cm and DB = 3 cm, then value of x is
 △ABC,DE||BC. If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1)

Updated On: May 31, 2025
  • 1
  • \(\frac{1}{2}\)
  • --1
  • \(\frac{1}{3}\)
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The Correct Option is B

Solution and Explanation

Given:
In \(\triangle ABC\), \(DE \parallel BC\).
Lengths: \(AE = (2x + 1) \, \text{cm}\), \(EC = 4 \, \text{cm}\), \(AD = (x + 1) \, \text{cm}\), \(DB = 3 \, \text{cm}\).
Find value of \(x\).

Step 1: Use Basic Proportionality Theorem (Thales theorem)
Since \(DE \parallel BC\),
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Step 2: Substitute given values
\[ \frac{x + 1}{3} = \frac{2x + 1}{4} \]

Step 3: Cross multiply and solve for \(x\)
\[ 4(x + 1) = 3(2x + 1) \] \[ 4x + 4 = 6x + 3 \] \[ 4 - 3 = 6x - 4x \] \[ 1 = 2x \implies x = \frac{1}{2} \]

Final Answer:
\[ \boxed{\frac{1}{2}} \]
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