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Triangles
NCERT Exemplar
CBSE
Grade 10

Question:

In $\Delta ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 4x - 3, AE = 8x - 7, BD = 3x - 1$ and $CE = 5x - 3$, find the value of $x$.

Step-by-Step Solution

Key Concept: BPT: $\dfrac{AD}{BD} = \dfrac{AE}{CE}$.
$\dfrac{4x - 3}{3x - 1} = \dfrac{8x - 7}{5x - 3} \Rightarrow (4x - 3)(5x - 3) = (8x - 7)(3x - 1)$. [0.5 Mark]
$20x^2 - 12x - 15x + 9 = 24x^2 - 8x - 21x + 7 \Rightarrow 20x^2 - 27x + 9 = 24x^2 - 29x + 7$. [0.5 Mark]
$4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0 \Rightarrow (2x + 1)(x - 1) = 0$. [0.5 Mark]
$x = 1$ or $x = -1/2$. Since lengths must be positive, $x = 1$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
BPT equation setup: 0.5 Mark
Expanding cross-product: 0.5 Mark
Solving quadratic for $x = 1$: 0.5 Mark
Rejecting negative root and concluding $x = 1$: 0.5 Mark

Correct Answer:
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