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.
- ½ mark
Since , corresponding sides are proportional: .
- 1 mark
CM and RN are medians, so and . Hence
- 1½ marks
Also [corresponding angles of similar triangles and ]. So in and , and included ; therefore [SAS similarity].
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
- Chapter: Triangles (notes, formulas, where marks are lost and every past question)
- Topic: Problems on Similar Triangles
More questions from Triangles
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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.