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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 1 · Q35
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\displaystyle \Delta \mathrm{PQR}$. Show that $\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E . Prove that $\displaystyle \mathrm{EL}=2 \mathrm{BL}$.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\displaystyle \Delta \mathrm{PQR}$. Show that $\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E . Prove that $\displaystyle \mathrm{EL}=2 \mathrm{BL}$.
Marking-scheme solution
In \(\displaystyle \Delta \mathrm{ABC}\) and \(\displaystyle \Delta \mathrm{PQR}\)
\[\begin{aligned}
& \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M} \\
& \frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M}
\end{aligned}
\]
( ∵ D is midpoint of BC and M is midpoint of QR)
\[\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM} \\
& \Rightarrow \angle \mathrm{~B}=\angle \mathrm{Q}-\text { (i) }
\end{aligned}
\]
Now, In \(\displaystyle \triangle \mathrm{ABC}\) and \(\displaystyle \triangle \mathrm{PQR}\)
\[\begin{array}{ll}
\frac{A B}{P Q}=\frac{B C}{Q R} & \text { (given) } \\
\angle \mathrm{B}=\angle \mathrm{Q} & \text { from (i) } \\
\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} &
\end{array}
\]
In \(\displaystyle \Delta \mathrm{BMC}\) and \(\displaystyle \Delta \mathrm{EMD}\)
\[\begin{aligned}
& \mathrm{MC}=\mathrm{MD} \\
& \angle \mathrm{CMB}=\angle \mathrm{EMD} \\
& \angle \mathrm{MBC}=\angle \mathrm{MED} \\
& \therefore \triangle \mathrm{BMC} \cong \triangle \mathrm{EMD} \\
& \Rightarrow \mathrm{BC}=\mathrm{DE}
\end{aligned}
\]
But AD = BC
\[\therefore \mathrm{AD}=\mathrm{DE}
\]
\[\Rightarrow \mathrm{AE}=2 \mathrm{BC}
\]
\(\displaystyle \Delta \mathrm{AEL} \sim \Delta \mathrm{CBL}\)
\[\therefore \frac{E L}{B L}=\frac{A E}{B C}
\]
\[\Longrightarrow \frac{E L}{B L}=\frac{2 B C}{B C}
\]
\[\Longrightarrow \frac{E L}{B L}=2
\]
\[\Rightarrow \mathrm{EL}=2 \mathrm{BL}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.