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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 4 · Set 2 · Q35
In the given figure, $\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BCA}$; prove that $\displaystyle \Delta \mathrm{ACB} \sim \Delta \mathrm{ADC}$. Hence find BD if $\displaystyle \mathrm{AC}=8 \mathrm{~cm}$ and $\displaystyle \mathrm{AD}=3 \mathrm{~cm}$.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
In the given figure, $\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BCA}$; prove that $\displaystyle \Delta \mathrm{ACB} \sim \Delta \mathrm{ADC}$. Hence find BD if $\displaystyle \mathrm{AC}=8 \mathrm{~cm}$ and $\displaystyle \mathrm{AD}=3 \mathrm{~cm}$.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Marking-scheme solution
In \(\displaystyle \Delta \mathrm{ACB}\) and \(\displaystyle \Delta \mathrm{ADC}\),
\[\begin{aligned}
& \angle \mathrm{ACB}=\angle \mathrm{ADC} \\
& \angle \mathrm{~A}=\angle \mathrm{A} \\
& \therefore \Delta \mathrm{ACB} \sim \Delta \mathrm{ADC} \\
& \therefore \frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AC}} \Rightarrow \frac{8}{3}=\frac{\mathrm{AB}}{8} \\
& \Rightarrow \mathrm{AB}=\frac{64}{3} \\
& \mathrm{AB}-\mathrm{AD}=\frac{64}{3}-3=\frac{55}{3} \mathrm{~cm}
\end{aligned}
\]
Correct given, to prove, figure and construction
Correct Proof
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.