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Triangles
RD Sharma
CBSE
Grade 10
Question:
In trapezium $ABCD$ with $AB \parallel DC$, diagonals $AC$ and $BD$ intersect at $O$. If $AO = 3\text{ cm}$, $OC = x - 3$, $BO = 3x - 19$ and $OD = x - 5$, then $x$ is:
(a) $8$ or $9$
(b) $7$ or $8$
(c) $8$ or $10$
(d) $9$ or $10$
(a) $8$ or $9$
(b) $7$ or $8$
(c) $8$ or $10$
(d) $9$ or $10$
Step-by-Step Solution
Key Concept: $\Delta AOB \sim \Delta COD \Rightarrow \dfrac{AO}{OC} = \dfrac{BO}{OD} \Rightarrow \dfrac{3}{x-3} = \dfrac{3x-19}{x-5} \Rightarrow 3(x-5) = (x-3)(3x-19) \Rightarrow 3x - 15 = 3x^2 - 28x + 57 \Rightarrow 3x^2 - 31x + 72 = 0 \Rightarrow (3x - 8)(x - 9) = 0 \Rightarrow x = 8$ or $9$.
$\dfrac{3}{x-3} = \dfrac{3x-19}{x-5} \Rightarrow 3x - 15 = 3x^2 - 28x + 57 \Rightarrow 3x^2 - 31x + 72 = 0$. [0.5 Mark]
$(3x - 8)(x - 9) = 0$. For integer values, $x = 8$ or $x = 9$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Forming quadratic $3x^2 - 31x + 72 = 0$: 0.5 Mark
Solving $x = 8$ or $9$: 0.5 Mark
$(3x - 8)(x - 9) = 0$. For integer values, $x = 8$ or $x = 9$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Forming quadratic $3x^2 - 31x + 72 = 0$: 0.5 Mark
Solving $x = 8$ or $9$: 0.5 Mark
Correct Answer: $8$ or $9$
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