CBSE 2025 · Region 3 · Set 1 · Q34 · 5 marks
In the given figure, $\displaystyle \mathrm{PA}, \mathrm{QB}$ and RC are perpendicular to AC. If $\displaystyle \mathrm{PA}=\mathrm{x}$ units, $\displaystyle \mathrm{QB}=\mathrm{y}$ units and $\displaystyle \mathrm{RC}=\mathrm{z}$ units, prove that $\displaystyle \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{z}}=\frac{1}{\mathrm{y}}$.
Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $\displaystyle \triangle \mathrm{PQR}$. Show that $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.
In the given figure, $\displaystyle \mathrm{PA}, \mathrm{QB}$ and RC are perpendicular to AC. If $\displaystyle \mathrm{PA}=\mathrm{x}$ units, $\displaystyle \mathrm{QB}=\mathrm{y}$ units and $\displaystyle \mathrm{RC}=\mathrm{z}$ units, prove that $\displaystyle \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{z}}=\frac{1}{\mathrm{y}}$.
Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $\displaystyle \triangle \mathrm{PQR}$. Show that $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.
Marking-scheme solution
\(\displaystyle \Delta \mathrm{ABQ} \sim \Delta \mathrm{ACR}\)
\(\displaystyle \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{QB}}{\mathrm{RC}}=\frac{\mathrm{y}}{\mathrm{z}}\)
Similarly, \(\displaystyle \Delta \mathrm{CBQ} \sim \Delta \mathrm{CAP}\)
\(\displaystyle \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{QB}}{\mathrm{PA}}=\frac{\mathrm{y}}{\mathrm{x}}\)
On adding (i) & (ii), we get
\[\begin{aligned}
& \frac{A B}{A C}+\frac{B C}{A C}=\frac{y}{z}+\frac{y}{x} \\
& \frac{A B+B C}{A C}=y\left(\frac{1}{z}+\frac{1}{x}\right) \\
& \frac{A C}{A C}=y\left(\frac{1}{z}+\frac{1}{x}\right) \\
& \therefore \frac{1}{x}+\frac{1}{z}=\frac{1}{y}
\end{aligned}
\]
(b)
In \(\displaystyle \triangle \mathrm{ABD}\) and \(\displaystyle \triangle \mathrm{PQM}\)
\[\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \text { (given) } \\
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\
& \therefore \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM} \\
& \therefore \angle \mathrm{~B}=\angle \mathrm{Q} \\
& \mathrm{In} \triangle \mathrm{ABC} \text { and } \triangle \mathrm{PQR} \\
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}} \text { and } \angle \mathrm{B}=\angle \mathrm{Q} \\
& \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}
\end{aligned}
\]
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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.