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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 1 · Set 1 · Q30
(A) In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that ∠ POQ =90^°.
In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $\displaystyle \angle \mathrm{POQ}=90^{\circ}$.
A circle with centre O and radius $\displaystyle 8$ cm is inscribed in a quadrilateral ABCD in which $\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ are the points of contact as shown. If AD is perpendicular to $\displaystyle \mathrm{DC}, \mathrm{BC}=30 \mathrm{~cm}$ and $\displaystyle \mathrm{BS}=24 \mathrm{~cm}$, then find the length DC.
Marking-scheme solution
Join OR.
\[\triangle \mathrm{AOQ} \cong \triangle \mathrm{ROQ} \Rightarrow \angle \mathrm{AOQ}=\angle \mathrm{ROQ}
\]
\[\triangle \mathrm{BOP} \cong \triangle \mathrm{ROP} \Rightarrow \angle \mathrm{BOP}=\angle \mathrm{ROP}
\]
Since \(\displaystyle \angle \mathrm{AOR}+\angle \mathrm{ROB}=180^{\circ}\)
\[\Rightarrow 2 \angle \mathrm{QOR}+2 \angle \mathrm{ROP}=180^{\circ}
\]
\[\Longrightarrow \angle \mathrm{QOR}+\angle \mathrm{ROP}=\angle \mathrm{POQ}=90^{\circ}
\]
Join OP and OQ.
\[\begin{aligned}
& \mathrm{BR}=\mathrm{BS}=24 \mathrm{~cm} \\
& \therefore \mathrm{CR}=6 \mathrm{~cm} \\
& \Rightarrow \mathrm{CQ}=6 \mathrm{~cm}
\end{aligned}
\]
Also, \(\displaystyle \mathrm{DQ}=\mathrm{OP}=8 \mathrm{~cm}\)
Hence, \(\displaystyle \mathrm{DC}=8+6=14 \mathrm{~cm}\)
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.