CBSE Class 10 Mathematics Standard 2024: question 34(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Figure: $△ABC$ with median AD (D midpoint of BC) and $△PQR$ with median PM (M midpoint of QR). Mark $\angle BAD=\angle 1$, $\angle DAC=\angle 2$, $\angle QPM=\angle 3$, $\angle MPR=\angle 4$. Given: $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$.

   1 mark

2. Construction: Produce AD to E such that AD = DE and join EC. Produce PM to L such that PM = ML and join LR.

   ½ mark

3. In $△ABD$ and $△ECD$: BD = DC, AD = DE, $\angle ADB=\angle EDC$ [vertically opposite]. So $△ABD\cong △ECD$ [SAS] $\Rightarrow AB=EC$ and $\angle BAD=\angle CED$ [CPCT]. Similarly, $△PQM\cong △LRM$, so $PQ=LR$ and $\angle QPM=\angle RLM$.

   1 mark

4. Now $\frac{EC}{LR}=\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}=\frac{2AD}{2PM}=\frac{AE}{PL}$. Hence $△AEC\sim △PLR$ [SSS similarity].

   1 mark

5. Therefore $\angle CAE=\angle RPL$, i.e. $\angle 2=\angle 4$, and $\angle AEC=\angle PLR$.

   ½ mark

6. Also $\angle 1=\angle BAD=\angle CED=\angle RLM=\angle QPM=\angle 3$. Adding, $\angle 1+\angle 2=\angle 3+\angle 4$, i.e. $\angle BAC=\angle QPR$.

   ½ mark

7. In $△ABC$ and $△PQR$: $\frac{AB}{PQ}=\frac{AC}{PR}$ and included $\angle BAC=\angle QPR$. Hence $△ABC\sim △PQR$ [SAS similarity].

   ½ mark

Answer: $△ABC\sim △PQR$ (by SAS similarity), hence proved.

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)