CBSE 2025 · Region 1 · Set 1 · Q24 · 2 marks
If $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ in which $\displaystyle \mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}, \mathrm{AC}=8 \mathrm{~cm}$ and $\displaystyle \mathrm{PR}=6 \mathrm{~cm}$, then find the length of ( $\displaystyle \mathrm{PQ}+\mathrm{QR}$ ).In the given figure, $\displaystyle \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$ and $\displaystyle \angle 1=\angle 2$, show that $\displaystyle \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}$.
If $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ in which $\displaystyle \mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}, \mathrm{AC}=8 \mathrm{~cm}$ and $\displaystyle \mathrm{PR}=6 \mathrm{~cm}$, then find the length of ( $\displaystyle \mathrm{PQ}+\mathrm{QR}$ ).
In the given figure, $\displaystyle \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$ and $\displaystyle \angle 1=\angle 2$, show that $\displaystyle \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}$.
Marking-scheme solution
\[\begin{aligned}
& \frac{6}{\mathrm{PQ}}=\frac{4}{\mathrm{QR}}=\frac{8}{6} \\
& \Rightarrow \mathrm{PQ}=\frac{9}{2} \mathrm{~cm} \text { or } 4.5 \mathrm{~cm}
\end{aligned}
\]
and \(\displaystyle \mathrm{QR}=3 \mathrm{~cm}\)
\[\therefore P Q+Q R=7.5 \mathrm{~cm}
\]
In \(\displaystyle \triangle \mathrm{PQR}, \angle 1=\angle 2 \therefore \mathrm{PR}=\mathrm{PQ}\)
\[\therefore \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}} \Rightarrow \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PQ}}
\]
Also, \(\displaystyle \angle 1=\angle 1\)
\[\therefore \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}
\]
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.