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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 6 · Set 1 · Q25
(A) In the given figure, ABC is a triangle in which DE ∥ BC. If AD =x, DB =x-2, AE =x+2 and EC =x-1, then find the value of x.
In the given figure, ABC is a triangle in which $\displaystyle \mathrm{DE} \| \mathrm{BC}$. If $\displaystyle \mathrm{AD}=x$, $\displaystyle \mathrm{DB}=x-2, \mathrm{AE}=x+2$ and $\displaystyle \mathrm{EC}=x-1$, then find the value of $\displaystyle x$.
Diagonals AC and BD of trapezium ABCD with AB $\displaystyle \| \mathrm{DC}$ intersect each other at point O . Show that $\displaystyle \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$.
Marking-scheme solution
In \(\displaystyle \triangle \mathrm{ABC}, \mathrm{DE} \| \mathrm{BC}\)
\[\begin{aligned}
& \therefore \frac{A D}{D B}=\frac{A E}{E C} \Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1} \\
& \mathrm{x}(\mathrm{x}-1)=(\mathrm{x}+2)(\mathrm{x}-2) \\
& \mathrm{x}^{2}-\mathrm{x}=\mathrm{x}^{2}-4 \Rightarrow \mathrm{x}=4
\end{aligned}
\]
In \(\displaystyle \Delta \mathrm{AOB}\) and \(\displaystyle \Delta \mathrm{COD}\),
\[\begin{aligned}
& \angle \mathrm{OAB}=\angle \mathrm{OCD} \\
& \angle \mathrm{OBA}=\angle \mathrm{ODC}
\end{aligned}
\]
Therefore, \(\displaystyle \Delta \mathrm{AOB} \sim \Delta \mathrm{COD}\)
\[\therefore \frac{O A}{O C}=\frac{O B}{O D}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.