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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 1 · Set 3 · Q34
(a) Sides AB and AC and median AM of a ABC are proportional to sides DE and DF and Median DN of another DEF. Show that ABC ∼ DEF. (b) ABCD is a…
(a)
Sides AB and AC and median AM of a $\displaystyle \triangle \mathrm{ABC}$ are proportional to sides DE and DF and Median DN of another $\displaystyle \triangle \mathrm{DEF}$. Show that $\displaystyle \triangle \mathrm{ABC}$ $\displaystyle \sim \triangle \mathrm{DEF}$.
(b)
ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that
(i)
$\displaystyle \frac{\mathrm{DP}}{\mathrm{PL}}=\frac{\mathrm{DC}}{\mathrm{BL}}$
(ii)
$\displaystyle \frac{\mathrm{DL}}{\mathrm{DP}}=\frac{\mathrm{AL}}{\mathrm{DC}}$
(iii)
If $\displaystyle \mathrm{LP}: \mathrm{PD}=2: 3$ then find $\displaystyle \mathrm{BP}: \mathrm{BC}$.
Marking-scheme solution
Extend AM to A' so that AM = A'M and DN to D' so that \(\displaystyle \mathrm{DN}=\mathrm{D}^{\prime} \mathrm{N}\). Join A'C and D'F.
\(\displaystyle \Delta \mathrm{AMB} \sim \Delta \mathrm{A}^{\prime} \mathrm{MC}\)
\(\displaystyle \Rightarrow \mathrm{AB}=\mathrm{A}^{\prime} \mathrm{C}\).
Similarly, \(\displaystyle \mathrm{DE}=\mathrm{D}{ }^{\prime} \mathrm{F}\)
Given \(\displaystyle \frac{A B}{D E}=\frac{A C}{D F}=\frac{A M}{D N}\)
\(\displaystyle \Rightarrow \frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{D}^{\prime} \mathrm{F}}=\frac{\mathrm{AA}^{\prime} / 2}{\mathrm{DD}^{\prime} / 2}\)
\(\displaystyle \therefore \triangle \mathrm{AA}^{\prime} \mathrm{C} \sim \triangle \mathrm{DDD}^{\prime} \mathrm{F}\)
\(\displaystyle \therefore \angle 1=\angle 2\)
Similarly, \(\displaystyle \angle 3=\angle 4\)
\(\displaystyle \Rightarrow \angle 1+\angle 3=\angle \mathrm{A}=\angle 2+\angle 4=\angle \mathrm{D}\)
Hence \(\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}(\mathrm{SAS})\)
(i)
\(\displaystyle \Delta \mathrm{DPC} \sim \Delta \mathrm{LPB}\)
\[\Rightarrow \frac{D P}{P L}=\frac{P C}{P B}=\frac{D C}{B L} \text { - (i) }
\]
(ii) As BC || AD
\[\therefore \Delta \mathrm{LPB} \sim \Delta \mathrm{LDA}
\]
In \(\displaystyle \triangle \mathrm{DLA}, \mathrm{AD} 11 \mathrm{BP}\)
\[\Rightarrow \frac{L P}{D P}=\frac{L B}{A B}
\]
\[\Rightarrow \frac{L P}{D P}+1=\frac{L B}{A B}+1
\]
\[\Rightarrow \frac{D L}{D P}=\frac{A L}{A B}
\]
\[\Rightarrow \frac{D L}{D P}=\frac{A L}{C D}(\mathrm{AB}=\mathrm{CD})
\]
(iii) \(\displaystyle \frac{L P}{L D}=\frac{P B}{A D}(\Delta \mathrm{LPB} \sim \Delta \mathrm{LDA})\)
\[\Rightarrow \frac{2}{5}=\frac{P B}{B C}(\mathrm{AD}=\mathrm{BC})
\]
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.