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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 1 · Q34
In the given figure, $\displaystyle \Delta \mathrm{FEC} \cong \triangle \mathrm{GDB}$ and $\displaystyle \angle 1=\angle 2$. Prove that $\displaystyle \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$.
Sides AB and AC and median AD of a $\displaystyle \triangle \mathrm{ABC}$ are respectively proportional to sides PQ and PR and median PM of another $\displaystyle \triangle \mathrm{PQR}$. Show that $\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.
In the given figure, $\displaystyle \Delta \mathrm{FEC} \cong \triangle \mathrm{GDB}$ and $\displaystyle \angle 1=\angle 2$. Prove that $\displaystyle \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$.
Sides AB and AC and median AD of a $\displaystyle \triangle \mathrm{ABC}$ are respectively proportional to sides PQ and PR and median PM of another $\displaystyle \triangle \mathrm{PQR}$. Show that $\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.
Marking-scheme solution
\(\displaystyle \Delta \mathrm{FEC} \cong \Delta \mathrm{GDB}\)
Therefore, \(\displaystyle \angle 3=\angle 4\)
In \(\displaystyle \Delta \mathrm{ABC}\),
\[\angle 3=\angle 4
\]
\[\therefore \mathrm{AB}=\mathrm{AC}
\]
In \(\displaystyle \triangle \mathrm{ADE}, \quad \angle 1=\angle 2\)
\[A D=A E
\]
Dividing (ii) by (i)
\[\begin{aligned}
& \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}} \\
& \Rightarrow \mathrm{DE} \| \mathrm{BC} \\
& \angle 1=\angle 3 \text { and } \angle 2=\angle 4 \\
& \therefore \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}
\end{aligned}
\]
Produce AD to E such that \(\displaystyle \mathrm{AD}=\mathrm{DE}\) and join EC .
Produce PM to L such that PM = ML and join LR.
\[\begin{aligned}
& \therefore \Delta \mathrm{ABD} \cong \Delta \mathrm{ECD} \\
& \therefore \mathrm{AB}=\mathrm{EC}
\end{aligned}
\]
Similarly, \(\displaystyle \mathrm{PQ}=\mathrm{LR}\)
\[\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\
& \frac{\mathrm{EC}}{\mathrm{LR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}=\frac{\mathrm{AE}}{\mathrm{PL}} \\
& \therefore \triangle \mathrm{AEC} \sim \triangle \mathrm{PLR}
\end{aligned}
\]
\[\Rightarrow \angle 2=\angle 4
\]
Similarly, \(\displaystyle \angle 1=\angle 3\)
Adding both, \(\displaystyle \angle \mathrm{BAC}=\angle \mathrm{QPR}\)
\[\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.