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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 1 · Q29
(A) In the given figure, 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, $\displaystyle \triangle \mathrm{ABC}$ 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 $\displaystyle \triangle \mathrm{PQR}$ at S and extended sides PQ and PR at M and N respectively, then prove that : $\displaystyle \mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})$
Marking-scheme solution
\[\mathrm{AC}=\sqrt{3^{2}+4^{2}}=5 \mathrm{~cm}
\]
Let \(\displaystyle \mathrm{BE}=\mathrm{BD}=x \mathrm{~cm}\)
\[\mathrm{AD}=4-x=\mathrm{AF}, \mathrm{CE}=3-x=\mathrm{CF}
\]
\(\displaystyle \mathrm{AF}+\mathrm{CF}=\mathrm{AC} \Rightarrow 4-x+3-x=5\)
\(\displaystyle \therefore x=1\)
\(\displaystyle \mathrm{BD}=\mathrm{BE}=1\) and \(\displaystyle \angle \mathrm{B}=90^{\circ}\)
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.
\[\left.\begin{array}{l}
\mathrm{PM}=\mathrm{PN} \\
\mathrm{QS}=\mathrm{QM} \\
\mathrm{RS}=\mathrm{RN}
\end{array}\right\}
\]
\[\begin{aligned}
\mathrm{PM}+\mathrm{PN} & =\mathrm{PQ}+\mathrm{QM}+\mathrm{PR}+\mathrm{RN} \\
2 \mathrm{PM} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{PR}+\mathrm{RS} \\
& =\mathrm{PQ}+\mathrm{QS}+\mathrm{RS}+\mathrm{PR}
\end{aligned}
\]
\[=\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}
\]
\[\therefore P M=\frac{1}{2}(P Q+Q R+P R)
\]
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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.