Skip to content
MathsCBSE Class 10 · Basic · Board exam 2023, set 430/6/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2023: question 28(b)

Set in the CBSE Class 10 board exam of 2023, paper 430/6/1, section C, question 28(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28(b) · 3 marks · Short answerTriangles

In the given figure, CM and RN are respectively the medians of ABC and PQR. If ABC PQR, then prove that AMC PNR.

Figure in the paper: Two triangles side by side. Triangle ABC on the left with base AMB (M between A and B) and apex C; segment CM is drawn as the median to AB. Triangle PQR on the right with base PNQ (N between P and Q) and apex R; segment RN is drawn as the median to PQ. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/6/1.

Hints

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

Hint 1

Write the ratio of sides from ΔABC ~ ΔPQR involving AB and AC.

Hint 2

M is the midpoint of AB, so AB = 2AM; combine with the equal included angle ∠A = ∠P.

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. Since , corresponding sides are proportional: .

    ½ mark
  2. CM and RN are medians, so and . Hence

    1 mark
  3. Also [corresponding angles of similar triangles and ]. So in and , and included ; therefore [SAS similarity].

    1½ marks

Answer: ΔAMC ~ ΔPNR by SAS similarity

Where the marks are

  • Writes AB/PQ = AC/PR from similarity (½ mark)
  • Uses medians: AB = 2AM, PQ = 2PN to get AM/PN = AC/PR (1 mark)
  • ∠A = ∠P and concludes SAS similarity (1½ marks)

Revise the chapter

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.