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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 2 · Q28
(A) In the given figure, Δ ABC is a right triangle in which ∠ B =90^°, AB =4 cm and BC =3 cm. Find the radius of the circle inscribed in the triangle…
In the given figure, $\displaystyle \Delta \mathrm{ABC}$ is a right triangle in which $\displaystyle \angle \mathrm{B}=90^{\circ}, \mathrm{AB}=4 \mathrm{~cm}$ and $\displaystyle \mathrm{BC}=3 \mathrm{~cm}$. Find the radius of the circle inscribed in the triangle ABC.
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
\[\begin{array}{l}
\mathrm{AC}=\sqrt{3^{2}+4^{2}}=5 \mathrm{~cm}
\text { Let } \mathrm{BE}=\mathrm{BD}=x \mathrm{~cm}
\mathrm{AD}=4-x=\mathrm{AF}, \mathrm{CE}=3-x=\mathrm{CF}
\mathrm{AF}+\mathrm{CF}=\mathrm{AC} \Rightarrow 4-x+3-x=5
\therefore x=1
\mathrm{BD}=\mathrm{BE}=1 \text { and } \angle \mathrm{B}=90^{\circ}
\end{array}
\]
Hence radius of circle $\displaystyle =x=1 \mathrm{~cm}$
ALTERNATE SOLUTION:
\[\mathrm{AC}=\sqrt{3^{2}+4^{2}}=5 \mathrm{~cm}
\]
Let $\displaystyle r$ be the radius of the circle
\[\operatorname{ar}(\triangle A B C)=\frac{1}{2} \times 4 \times 3=6 \mathrm{~cm}^{2}
\]
Also, $\displaystyle \operatorname{ar}(\triangle A B C)=\left(\frac{1}{2} \times r \times 4\right)+\left(\frac{1}{2} \times r \times 3\right)+\left(\frac{1}{2} \times r \times 5\right)$
\[\Rightarrow 6 r=6 \Rightarrow r=1
\]
Hence the radius of the circle is $\displaystyle 1$ cm.
\[\mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})
\]
\[\begin{array}{l}
\mathrm{PM}=\mathrm{PN}
\mathrm{QS}=\mathrm{QM}
\mathrm{RS}=\mathrm{RN}
\mathrm{PM}+\mathrm{PN}=\mathrm{PQ}+\mathrm{QM}+\mathrm{PR}+\mathrm{RN}
\qquad \begin{aligned}
2 \mathrm{PM} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{PR}+\mathrm{RS}
& =\mathrm{PQ}+\mathrm{QS}+\mathrm{RS}+\mathrm{PR}
\end{aligned}
\end{array}
\]
$\displaystyle \begin{array}{c}=\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}
\therefore \mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})\end{array}$
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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.