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Class 10th Mathematics Chapter 6 Triangles – Important Questions

Important questions 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 important questions for Class 10th Mathematics Chapter 6 Triangles?

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

Are Triangles important questions free?

Yes, Notebook gives free NCERT CBSE important 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 important questions for quick revision?

You are on the right page: this important questions 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 PYQ

Triangles

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1 Mark
State the Basic Proportionality Theorem.
Ans: If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
2 Marks
In △ABC, DE ∥ BC. If AD = 4 cm, DB = 6 cm, AE = 3 cm, find EC.
Ans: By BPT: AD/DB = AE/EC → 4/6 = 3/EC → EC = 3×6/4 = 4.5 cm.
3 Marks
Prove that the ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides.
Ans: Let △ABC ~ △PQR. Draw AM ⊥ BC and PN ⊥ QR. Area(ABC)/Area(PQR) = (½×BC×AM)/(½×QR×PN). In △ABM and △PQN: ∠B=∠Q (similar triangles), ∠M=∠N=90° → △ABM~△PQN → AM/PN = AB/PQ. Also AB/PQ = BC/QR. So ratio of areas = (BC/QR)×(AB/PQ) = (AB/PQ)² = (BC/QR)².
5 Marks
In a right triangle, prove that the square of the hypotenuse equals the sum of squares of the other two sides (Pythagoras Theorem).
Ans: Given: △ABC right-angled at B. Draw BD ⊥ AC. △ADB ~ △ABC → AD/AB = AB/AC → AB² = AD×AC. △BDC ~ △ABC → DC/BC = BC/AC → BC² = DC×AC. Adding: AB²+BC² = AC(AD+DC) = AC×AC = AC². Hence proved.
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