Question

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

• A. \(1\)
• B. \(\frac{1}{3}\)
• C. \(-1\)
• D. \(\frac{1}{2}\)

Hint:

Use the Basic Proportionality Theorem for lines parallel to one side in a triangle: \(\frac{AD}{DB} = \frac{AE}{EC}\).

Answer 1

The Correct Option is D

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

To find:
The value of \(x\).

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

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

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

 

Summary:
\[ \boxed{x = \frac{1}{2}} \]