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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 6 · Set 3 · Q33
(A) Prove that a parallelogram circumscribing a circle is a rhombus.
Prove that a parallelogram circumscribing a circle is a rhombus.
In the given figure, tangents PQ and PR are drawn to a circle such that $\displaystyle \angle \mathrm{RPQ}=30^{\circ}$. A chord RS is drawn parallel to the tangent PQ . Find the measure of $\displaystyle \angle \mathrm{RQS}$.
Marking-scheme solution
ABCD is a parallelogram touching the circle at \(\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}\) by sides AB, BC, CD, DA respectively.
We know that tangents drawn from the external point to a circle are equal.
\[\therefore \mathrm{AP}=\mathrm{AS} \quad-------(\mathrm{i})
\]
\[\begin{aligned}
\mathrm{PB}=\mathrm{BQ} & \\
\mathrm{CR}=\mathrm{CQ} & \\
\mathrm{DR}=\mathrm{DS} & \\
& \text { Adding (i), (ii), (iii), (iv) } \\
& (\mathrm{AP}+\mathrm{PB})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\
& \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \\
& \mathrm{ABCD} \text { is a parallelogram } \\
& \Rightarrow \mathrm{AB}=\mathrm{CD}, \mathrm{AD}=\mathrm{BC} \\
& \Rightarrow 2 \mathrm{AB}=2 \mathrm{AD} \Rightarrow \mathrm{AB}=\mathrm{AD} \\
& \Rightarrow \mathrm{ABCD} \text { is a rhombus. }
\end{aligned}
\]
\[\begin{aligned}
& (\mathrm{AP}+\mathrm{PB})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\
& \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \\
& \mathrm{ABCD} \text { is a parallelogram } \\
& \Rightarrow \mathrm{AB}=\mathrm{CD}, \mathrm{AD}=\mathrm{BC} \\
& \Rightarrow 2 \mathrm{AB}=2 \mathrm{AD} \Rightarrow \mathrm{AB}=\mathrm{AD} \\
& \Rightarrow \mathrm{ABCD} \text { is a rhombus. }
\end{aligned}
\]
PQ = PR (tangents drawn from an external point to the circle )
\[\therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ}
\]
In \(\displaystyle \triangle \mathrm{PQR}, \angle \mathrm{PQR}=\angle \mathrm{PRQ}=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}\)
Draw a perpendicular QL from Q to QP
Now, \(\displaystyle \angle \mathrm{PQL}=90^{\circ}\)
\(\displaystyle \therefore \angle \mathrm{RQL}=90^{\circ}-75^{\circ}=15^{\circ}\)
\(\displaystyle \Delta \mathrm{RQL} \cong \Delta \mathrm{SQL}(\mathrm{SAS})\)
\(\displaystyle \therefore \angle \mathrm{RQL}=\angle \mathrm{SQL}=15^{\circ}\)
\(\displaystyle \therefore \angle \mathrm{RQS}=15^{\circ}+15^{\circ}=30^{\circ}\)
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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.