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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 2 · Q28 (B)
In the given figure, if a circle touches the side QR of Δ PQR at S and extended sides PQ and PR at M and N respectively, then prove that: PM =1/2( PQ…
In the given figure, if a circle touches the side QR of $\displaystyle \Delta \mathrm{PQR}$ at S and extended sides PQ and PR at M and N respectively, then prove that : \[\mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}) \]
Marking-scheme solution
\[\mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})
\]
\[\begin{array}{l}
\mathrm{PM}=\mathrm{PN}
\mathrm{QS}=\mathrm{QM}
\mathrm{RS}=\mathrm{RN}
\mathrm{PM}+\mathrm{PN}=\mathrm{PQ}+\mathrm{QM}+\mathrm{PR}+\mathrm{RN}
\qquad \begin{aligned}
2 \mathrm{PM} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{PR}+\mathrm{RS}
& =\mathrm{PQ}+\mathrm{QS}+\mathrm{RS}+\mathrm{PR}
\end{aligned}
\end{array}
\]
$\displaystyle \begin{array}{c}=\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}
\therefore \mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})\end{array}$
\[\begin{array}{l}
\mathrm{PM}=\mathrm{PN}
\mathrm{QS}=\mathrm{QM}
\mathrm{RS}=\mathrm{RN}
\mathrm{PM}+\mathrm{PN}=\mathrm{PQ}+\mathrm{QM}+\mathrm{PR}+\mathrm{RN}
\qquad \begin{aligned}
2 \mathrm{PM} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{PR}+\mathrm{RS}
& =\mathrm{PQ}+\mathrm{QS}+\mathrm{RS}+\mathrm{PR}
\end{aligned}
\end{array}
\]
$\displaystyle \begin{array}{c}=\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}
\therefore \mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})\end{array}$More from Circles
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.