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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 4 · Set 3 · Q32
In a $\displaystyle \triangle \mathrm{PQR}, \mathrm{N}$ is a point on PR, such that $\displaystyle \mathrm{QN} \perp \mathrm{PR}$. If $\displaystyle \mathrm{PN} \times \mathrm{NR}=\mathrm{QN}^{2}$, prove that $\displaystyle \angle \mathrm{PQR}=90^{\circ}$.In the given figure, $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \Delta \mathrm{DBC}$ are on the same base BC. If AD intersects BC at O, prove that $\displaystyle \frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}$
In a $\displaystyle \triangle \mathrm{PQR}, \mathrm{N}$ is a point on PR, such that $\displaystyle \mathrm{QN} \perp \mathrm{PR}$. If $\displaystyle \mathrm{PN} \times \mathrm{NR}=\mathrm{QN}^{2}$, prove that $\displaystyle \angle \mathrm{PQR}=90^{\circ}$.
In the given figure, $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \Delta \mathrm{DBC}$ are on the same base BC. If AD intersects BC at O, prove that $\displaystyle \frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}$
Marking-scheme solution
(A)
\(\displaystyle \mathrm{PN} \times \mathrm{NR}=\mathrm{QN}^{2}\)
\(\displaystyle \frac{\mathrm{PN}}{\mathrm{QN}}=\frac{\mathrm{QN}}{\mathrm{NR}}\)
\[\begin{aligned}
& \angle \mathrm{PNQ}=\angle \mathrm{QNR} \\
& \Delta \mathrm{PNQ} \sim \Delta \mathrm{QNR} \\
& \Rightarrow \angle 2=\angle \mathrm{P} \text { and } \angle 1=\angle \mathrm{R} \\
& \Rightarrow \angle 1+\angle 2=\angle \mathrm{P}+\angle \mathrm{R} \\
& \Rightarrow \angle \mathrm{PQR}=\angle \mathrm{P}+\angle \mathrm{R} \\
& \text { In } \triangle \mathrm{PQR}, \angle \mathrm{P}+\angle \mathrm{PQR}+\angle \mathrm{R}=180^{\circ} \\
& \Rightarrow 2 \angle \mathrm{PQR}=180^{\circ} \Rightarrow \angle \mathrm{PQR}=90^{\circ}
\end{aligned}
\]
(B)
Draw \(\displaystyle \mathrm{AL} \perp \mathrm{BC}\) and \(\displaystyle \mathrm{DM} \perp \mathrm{BC}\)
In \(\displaystyle \triangle A O L\) and \(\displaystyle \triangle D O M\),
\[\begin{aligned}
& \angle A O L=\angle D O M \\
& \angle A L O=\angle D M O
\end{aligned}
\]
\[\begin{aligned}
\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DBC})} & =\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AL}}{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{DM}} \\
& =\frac{A L}{D M}=\frac{A O}{D O}[\text { using (i) }]
\end{aligned}
\]
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