MathsCBSE Class 10 · Standard · Board exam 2024, set 30/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2024: question 33(b)

Set in the CBSE Class 10 board exam of 2024, paper 30/2/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerTriangles

Sides $A B$ and $A C$ and median $A D$ to $△ A B C$ are respectively proportional to sides $P Q$ and $P R$ and median $P M$ of another triangle $P QR$ . Show that $△ A B C \sim △ P QR$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/2/1.

Hints

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

Hint 1

Extend each median to twice its length beyond the mid-point and join to a vertex.

Hint 2

Show AB = EC by congruence; then compare triangles AEC and PNR using all three side ratios.

Hint 3

Once $∠ A = ∠ P$ , use SAS similarity.

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.

Figure: $△ A B C$ with median $A D$ ( $D$ mid-point of $B C$ ) and $△ P QR$ with median $P M$ ( $M$ mid-point of $QR$ ). Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ . Mark $∠ B A D = ∠3$ , $∠ D A C = ∠1$ , $∠ QP M = ∠4$ , $∠ M P R = ∠2$ .
1 mark
Construction: Produce $A D$ to $E$ so that $D E = A D$ and join $E C$ ; produce $P M$ to $N$ so that $M N = P M$ and join $N R$ .
½ mark
In $△ A D B$ and $△ E D C$ : $A D = E D$ , $B D = C D$ , $∠ A D B = ∠ E D C$ (vertically opposite). So $△ A D B ≅ △ E D C$ (SAS) $\Rightarrow A B = E C$ and $∠ B A D = ∠ C E D$ . Similarly $P Q = N R$ and $∠ QP M = ∠ R N M$ .
1 mark
Now $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} \Rightarrow \frac{E C}{N R} = \frac{A C}{P R} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ . Hence $△ A E C \sim △ P N R$ (SSS similarity).
1 mark
$\Rightarrow ∠ E A C = ∠ N P R$ , i.e. $∠1 = ∠2$ (corresponding angles of similar triangles).
½ mark
Also $∠ A E C = ∠ P N R$ ; since $∠3 = ∠ A E C$ and $∠4 = ∠ P N R$ , we get $∠3 = ∠4$ . Adding: $∠1 + ∠3 = ∠2 + ∠4$ , i.e. $∠ B A C = ∠ QP R$ .
½ mark
In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ A = ∠ P$ . Hence $△ A B C \sim △ P QR$ (SAS similarity).
½ mark

Answer: $△ A B C \sim △ P QR$ (SAS similarity) – proved

Where the marks are

Correct figure with given data (1 mark)
Construction: produce medians to double length and join EC, NR (½ mark)
$△ A D B ≅ △ E D C$ giving AB = EC (and PQ = NR) (1 mark)
$△ A E C \sim △ P N R$ by SSS (1 mark)
$∠1 = ∠2$ (½ mark)
$∠3 = ∠4$ and hence $∠ A = ∠ P$ (½ mark)
Concludes $△ A B C \sim △ P QR$ by SAS (½ mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/2/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2024: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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