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Triangles
NCERT Exemplar Ch 06
CBSE_NCERT_EXEMPLAR_CH06
Grade 10
Question:
In $\Delta ABC$, $DE \parallel BC$ such that $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$. The value of $x$ is:
$4$
$3$
$2$
$1$
$4$
$3$
$2$
$1$
Step-by-Step Solution
Key Concept: BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
Stepwise Solution:
$\dfrac{x}{x-2} = \dfrac{x+2}{x-1} \Rightarrow x(x-1) = (x-2)(x+2)$. [0.5 Mark]
$x^2 - x = x^2 - 4 \Rightarrow -x = -4 \Rightarrow x = 4$. [0.5 Mark]
Marking Scheme:
• BPT setup with algebraic terms: 0.5 Mark
• Solving for $x = 4$: 0.5 Mark
$\dfrac{x}{x-2} = \dfrac{x+2}{x-1} \Rightarrow x(x-1) = (x-2)(x+2)$. [0.5 Mark]
$x^2 - x = x^2 - 4 \Rightarrow -x = -4 \Rightarrow x = 4$. [0.5 Mark]
Marking Scheme:
• BPT setup with algebraic terms: 0.5 Mark
• Solving for $x = 4$: 0.5 Mark
Correct Answer: $4$
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