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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 1 · Set 2 · Q34
$\displaystyle \mathrm{PA}, \mathrm{QB}$ and RC are each perpendicular to AC . If $\displaystyle \mathrm{AP}=x, \mathrm{QB}=\mathrm{z}$, $\displaystyle \mathrm{RC}=\mathrm{y}, \mathrm{AB}=\mathrm{a}$ and $\displaystyle \mathrm{BC}=\mathrm{b}$, then prove that $\displaystyle \frac{1}{x}+\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{z}}$.
In the given figure, CD and RS are respectively the medians of $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \triangle \mathrm{PQR}$. If $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ then prove that :(i)$\displaystyle \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}$(ii)$\displaystyle \mathrm{AD} \times \mathrm{PR}=\mathrm{AC} \times \mathrm{PS}$
$\displaystyle \mathrm{PA}, \mathrm{QB}$ and RC are each perpendicular to AC . If $\displaystyle \mathrm{AP}=x, \mathrm{QB}=\mathrm{z}$, $\displaystyle \mathrm{RC}=\mathrm{y}, \mathrm{AB}=\mathrm{a}$ and $\displaystyle \mathrm{BC}=\mathrm{b}$, then prove that $\displaystyle \frac{1}{x}+\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{z}}$.
In the given figure, CD and RS are respectively the medians of $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \triangle \mathrm{PQR}$. If $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ then prove that :
(i)
$\displaystyle \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}$
(ii)
$\displaystyle \mathrm{AD} \times \mathrm{PR}=\mathrm{AC} \times \mathrm{PS}$
Marking-scheme solution
\(\displaystyle \Delta \mathrm{CQB} \sim \Delta \mathrm{CPA}\)
\[\begin{aligned}
& \Rightarrow \frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{x}}-\text { (i) } \\
& \text { Also } \Delta \mathrm{AQB} \sim \Delta \text { ARC } \\
& \Rightarrow \frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}=\frac{z}{y}-\text { (ii) } \\
& \text { from (i) and (ii) } \frac{\mathrm{z}}{\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{y}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}+\mathrm{b}}=1 \\
& \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{z}}
\end{aligned}
\]
\[\begin{aligned}
& \text { (i) } \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\
& \Rightarrow \angle \mathrm{~A}=\angle \mathrm{P} \\
& \text { and } \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \\
& \Rightarrow \frac{2 A D}{2 P S}=\frac{A C}{P R}
\end{aligned}
\]
\[\Rightarrow \frac{A D}{P S}=\frac{A C}{P R} \text { and } \angle \mathrm{A}=\angle \mathrm{P}
\]
Therefore \(\displaystyle \Delta \mathrm{ADC} \sim \Delta \mathrm{PSR}\)
\[\begin{aligned}
& \text { (ii)Hence } \frac{A D}{P S}=\frac{A C}{P R} \\
& \Rightarrow \mathrm{AD} \times \mathrm{PR}=\mathrm{AC} \times \mathrm{PS}
\end{aligned}
\]
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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.