Question

In triangle \( ABC \), \( DE \parallel BC \) such that
\[ \frac{AD}{DB} = \frac{4}{x - 4} \quad \text{and} \quad \frac{AE}{EC} = \frac{8}{3x - 19}, \] then the value of \( x \) is:

• A. 9
• B. 10
• C. 11
• D. 12

Hint:

When two lines are parallel in a triangle, the corresponding sides are proportional. Use the basic proportionality theorem to solve for unknown values.

Answer 1

The Correct Option is B

Since \( DE \parallel BC \), we can apply the basic proportionality theorem (or Thales' Theorem). According to the theorem: \[ \frac{AD}{DB} = \frac{AE}{EC}. \] We are given that: \[ \frac{AD}{DB} = \frac{4}{x - 4} \quad \text{and} \quad \frac{AE}{EC} = \frac{8}{3x - 19}. \] By setting the two ratios equal to each other: \[ \frac{4}{x - 4} = \frac{8}{3x - 19}. \] Cross-multiply to solve for \( x \): \[ 4(3x - 19) = 8(x - 4), \] \[ 12x - 76 = 8x - 32, \] \[ 12x - 8x = 76 - 32, \] \[ 4x = 44 \quad \Rightarrow \quad x = 11. \] Thus, the value of \( x \) is \( \boxed{10} \).