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Class 10th Mathematics Chapter 6 Triangles – PYQ

Previous year questions (PYQ) with answers for Class 10th Mathematics Chapter 6 Triangles, based on the NCERT CBSE syllabus.

Topics covered in Triangles

Frequently asked questions

Where can I find previous year questions (PYQ) for Class 10th Mathematics Chapter 6 Triangles?

This page lists previous year questions with answers for Triangles. Topics covered: Basic proportionality theorem, Similarity of triangles, AA, SSS, SAS similarity, Pythagoras theorem.

Are Triangles PYQs free?

Yes, Notebook gives free previous year questions for Class 10 Mathematics.

How to prepare Triangles for Class 10 Mathematics exams?

Read the Triangles notes first, revise the key topics (Basic proportionality theorem, Similarity of triangles, AA, SSS, SAS similarity, Pythagoras theorem), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 10th Mathematics Chapter 6 Triangles pyq for quick revision?

You are on the right page: this pyq page for Triangles covers Class 10 Mathematics for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 10th Mathematics Chapter 6 Triangles important for the CBSE exam 2026-27?

Triangles is part of the Class 10 Mathematics NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 10th Mathematics Chapter 6 Triangles Notes · Class 10th Mathematics Chapter 6 Triangles Important Questions

Triangles

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3 Marks CBSE 2019
In △ABC, if AD is the median and E is the midpoint of AD, prove that ar(△BED) = ¼ ar(△ABC).
Ans: AD is median → ar(△ABD) = ½ ar(△ABC). E is midpoint of AD → BE is median of △ABD → ar(△BED) = ½ ar(△ABD) = ½ × ½ ar(△ABC) = ¼ ar(△ABC). Hence proved.
3 Marks CBSE 2021
Sides of triangles are given: (i) 7cm, 24cm, 25cm (ii) 3cm, 8cm, 6cm. Which is right-angled?
Ans: (i) 7²+24² = 49+576 = 625 = 25². Yes, right-angled. (ii) 3²+6² = 9+36 = 45 ≠ 64 = 8². Not right-angled.
5 Marks CBSE 2023
Prove that in a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.
Ans: Given △ABC, right angle at B. Draw BD ⊥ AC. △ADB ~ △ABC (AA) → AD/AB = AB/AC → AB² = AD·AC. △BDC ~ △ABC (AA) → CD/BC = BC/AC → BC² = CD·AC. Adding: AB²+BC² = AC(AD+CD) = AC·AC = AC². Hence proved.
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