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Tamil Nadu (TNBSE)Class 10 Mathematics★ 8-12 marks in the TN SSLC Class 10 Maths public exam through MCQs, theorem-based proofs and construction/problem questions

Geometry

Samacheer Kalvi (TN State Board) Class 10 Maths Chapter 4, Geometry. Similarity of triangles, the Basic Proportionality (Thales) theorem and the angle bisector theorem, the Pythagoras theorem and its converse, the concurrency theorems (Ceva and Menelaus), and tangents and secants of circles. Notes with theorems, worked examples, common mistakes, book-back practice and quick revision.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply similarity criteria for triangles
2Use the Thales and angle bisector theorems
3Apply the Pythagoras theorem and its converse
4State Ceva's and Menelaus' concurrency theorems
5Use tangent and secant properties of circles

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Why this chapter matters

Geometry develops reasoning through similarity and circle theorems. Thales, Pythagoras, the concurrency theorems and tangent properties are central to the TN SSLC exam, both as proofs and as problems.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 9 — Triangles and circles

Congruence and basic circle facts.

Geometry — Class 10 Maths (Samacheer Kalvi)

TN State Board (Samacheer Kalvi) Class 10 Mathematics, Chapter 4. Similar triangles, key theorems and circle geometry.

1. About this chapter

This chapter covers similarity, the Thales and angle bisector theorems, the Pythagoras theorem and its converse, the concurrency theorems (Ceva, Menelaus), and tangents/secants of circles.

2. Similar triangles

Two triangles are similar if their corresponding angles are equal and corresponding sides are proportional.
Criteria: AA, SSS, SAS similarity.
In similar triangles, the ratio of areas = (ratio of corresponding sides)².

3. Key theorems

Basic Proportionality Theorem (Thales): a line drawn parallel to one side of a triangle divides the other two sides in the same ratio — AD/DB = AE/EC.
Angle Bisector Theorem: the internal bisector of an angle divides the opposite side in the ratio of the other two sides — BD/DC = AB/AC.
Pythagoras theorem: in a right triangle, hypotenuse² = base² + height²; its converse is also true (used to test right angles).

4. Concurrency and circles

Ceva's theorem: cevians AD, BE, CF are concurrent iff (BD/DC)(CE/EA)(AF/FB) = 1.
Menelaus' theorem: points on the sides are collinear iff (BD/DC)(CE/EA)(AF/FB) = −1 (signed).
Tangent: a line touching a circle at one point; it is perpendicular to the radius at the point of contact. Lengths of the two tangents drawn from an external point are equal.

5. Worked examples

Example 1. In a right triangle, base = 6 cm, height = 8 cm. Find the hypotenuse. h² = 6² + 8² = 36 + 64 = 100 → h = 10 cm.

Example 2. A line parallel to BC meets AB at D and AC at E with AD = 2, DB = 3, AE = 4. Find EC. By Thales: AD/DB = AE/EC → 2/3 = 4/EC → EC = 6.

Example 3. Two tangents are drawn from a point 13 cm from the centre of a circle of radius 5 cm. Find the tangent length. Tangent² = 13² − 5² = 169 − 25 = 144 → length = 12 cm.

6. Common mistakes

Mistake: Saying area ratio = side ratio in similar triangles. Fix: Area ratio = (side ratio)².
Mistake: Using Pythagoras in a non-right triangle. Fix: It applies only to right-angled triangles.
Mistake: Forgetting tangent ⟂ radius. Fix: A tangent is perpendicular to the radius at the point of contact.

7. Practice (book-back style)

State the Basic Proportionality (Thales) theorem.
State the Pythagoras theorem.
The sides of two similar triangles are in ratio 2 : 3. Find the ratio of their areas.
From a point 10 cm from the centre of a circle of radius 6 cm, find the tangent length.
State the angle bisector theorem.

8. Answer key

A line parallel to one side of a triangle divides the other two sides in the same ratio.
In a right triangle, hypotenuse² = base² + height².
(2 : 3)² = 4 : 9.
Tangent² = 10² − 6² = 64 → length = 8 cm.
The internal bisector divides the opposite side in the ratio of the adjacent sides (BD/DC = AB/AC).

9. Quick revision

Chapter 4 · similarity, theorems, circles.
Similar triangles: AA/SSS/SAS; area ratio = (side ratio)².
Thales: AD/DB = AE/EC; angle bisector: BD/DC = AB/AC.
Pythagoras: h² = b² + p² (and converse).
Ceva = 1, Menelaus = −1; tangent ⟂ radius; equal tangents from a point.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Thales theorem

AD/DB = AE/EC

Line parallel to one side.

Pythagoras theorem

hypotenuse² = base² + height²

Right triangle only; converse holds.

Area ratio of similar triangles

(ratio of sides)²

Not the side ratio itself.

Ceva / Menelaus

product = 1 (Ceva) / −1 (Menelaus)

Concurrency / collinearity.

Tangent from external point

tangent² = d² − r²

Tangent ⟂ radius; equal tangents.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying area ratio = side ratio in similar triangles

✓ Area ratio = (side ratio)².

WATCH OUT

✗ Using Pythagoras in a non-right triangle

✓ It applies only to right-angled triangles.

WATCH OUT

✗ Forgetting tangent ⟂ radius

✓ A tangent is perpendicular to the radius at the point of contact.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Multiple choice questions

Questions

Theorem / proof questions

Thales, Pythagoras and angle bisector proofs.

Questions

Problem solving

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Concept

State the Basic Proportionality (Thales) theorem.

▶ Show solution

A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

Q2EASY· Numerical

In a right triangle, base = 6 cm and height = 8 cm. Find the hypotenuse.

▶ Show solution

h² = 6² + 8² = 100 → h = 10 cm.

Q3EASY· Concept

The sides of two similar triangles are in ratio 2 : 3. Find the ratio of their areas.

▶ Show solution

(2 : 3)² = 4 : 9.

Q4MEDIUM· Numerical

From a point 13 cm from the centre of a circle of radius 5 cm, find the tangent length.

▶ Show solution

Tangent² = 13² − 5² = 144 → length = 12 cm.

Q5MEDIUM· Numerical

A line parallel to BC meets AB at D, AC at E with AD = 2, DB = 3, AE = 4. Find EC.

▶ Show solution

AD/DB = AE/EC → 2/3 = 4/EC → EC = 6.

Q6EASY· Concept

State the angle bisector theorem.

▶ Show solution

The internal bisector of an angle divides the opposite side in the ratio of the other two sides (BD/DC = AB/AC).

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Chapter 4 of Samacheer Kalvi Class 10 Mathematics.
•Similar triangles: AA/SSS/SAS; area ratio = (side ratio)².
•Thales: AD/DB = AE/EC; angle bisector: BD/DC = AB/AC.
•Pythagoras: h² = b² + p² and its converse.
•Ceva = 1 (concurrency), Menelaus = −1 (collinearity).
•Tangent ⟂ radius; tangent² = d² − r²; equal tangents from a point.

Tamil Nadu (TNBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-12 marks across MCQ, proofs and problems

Question type	Marks each	Typical count	What it tests
MCQ	1	1-2	Theorems and properties
Proof / Short	2-3	1-2	Thales, Pythagoras, angle bisector
Problem	2-5	1-2	Similarity and tangent lengths

Prep strategy

Learn the similarity criteria and area-ratio rule
Practise Thales and Pythagoras problems
Memorise Ceva = 1, Menelaus = −1
Use tangent² = d² − r² for tangent lengths

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Architecture

Similar triangles and Pythagoras are used in design and surveying.

Navigation

Tangents and distances apply to circular paths and ranges.

Engineering drawing

Proportionality underlies scaling of plans.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State the theorem before applying it
Square the side ratio for area problems
Draw clear, labelled figures
Use tangent² = d² − r² directly

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Prove the Thales theorem using areas.
Apply Ceva's theorem to show the medians are concurrent.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

TN SSLC Class 10 Public ExamHigh

Foundation / NTSE MathematicsMedium

School unit testsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

If the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled — so the converse is used to test for a right angle.

Both tangent lengths form right triangles with the same radius and the same distance to the centre, so by congruence the tangent lengths are equal.

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