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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q33
In the given figure, TP and TQ are tangents to a circle with centre M , touching another circle with centre N at A and B respectively. It is given that $\displaystyle \mathrm{MQ}=13 \mathrm{~cm}, \mathrm{NB}=8 \mathrm{~cm}, \mathrm{BQ}=35 \mathrm{~cm}$ and $\displaystyle \mathrm{TP}=80 \mathrm{~cm}$.(i)Name the quadrilateral MQBN.(ii)Is MN parallel to PA ? Justify your answer.(iii)Find length TB.(iv)Find length MN.
In the given figure, TP and TQ are tangents to a circle with centre M , touching another circle with centre N at A and B respectively. It is given that $\displaystyle \mathrm{MQ}=13 \mathrm{~cm}, \mathrm{NB}=8 \mathrm{~cm}, \mathrm{BQ}=35 \mathrm{~cm}$ and $\displaystyle \mathrm{TP}=80 \mathrm{~cm}$.
(i)
Name the quadrilateral MQBN.
(ii)
Is MN parallel to PA ? Justify your answer.
(iii)
Find length TB.
(iv)
Find length MN.
Marking-scheme solution
(i) \(\displaystyle \mathrm{BN} \| \mathrm{QM}\) and \(\displaystyle \mathrm{QB} \| \mathrm{MN}\) The quadrilateral MQBN is a
trapezium.
trapezium.(ii) No, as AN \(\displaystyle \neq \mathrm{PM}\)
(iii) \(\displaystyle \mathrm{TQ}=\mathrm{TP}=80 \mathrm{~cm}\)
\[\therefore \mathrm{TB}=80-35=45 \mathrm{~cm}
\]
(iv) \(\displaystyle \mathrm{MN}^{2}=\mathrm{NS}^{2}+\mathrm{MS}^{2}\)
\[\begin{aligned}
& =35^{2}+(13-8)^{2} \\
& =1225+25 \\
& =1250
\end{aligned}
\]
\[\therefore \mathrm{MN}=25 \sqrt{2} \mathrm{~cm}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.