CBSE Class 10 Mathematics Triangles Notes
About This Chapter
Chapter 6 - Triangles is one of the most comprehensive and high-weightage chapters in the CBSE Class 10 Mathematics syllabus. Building upon your earlier understanding of basic geometry, this chapter formally introduces the powerful concept of similarity - the idea that two figures can have the same shape but different sizes. This underpins much of advanced geometry and real-world measurement.
From architects scaling blueprints, to surveyors measuring distances across rivers, to engineers designing bridges - the principles of similar triangles and the Pythagoras theorem are at work everywhere. Understanding this chapter gives you the tools to prove geometric relationships rigorously and solve measurement problems that cannot be approached directly.
This chapter typically carries 10 to 12 marks in the CBSE board examination, making it one of the most important chapters for scoring well. Questions appear across all categories - MCQs, short answers, and long proof-based questions. Theorem proofs, especially BPT and the Pythagoras theorem, are directly asked in board papers.
What You Will Learn:
• The Basic Proportionality Theorem (Thales Theorem) and its converse, with full proofs
• Criteria for similarity of triangles: AA, SSS, and SAS similarity rules
• The relationship between areas of similar triangles and their corresponding sides
• Pythagoras theorem and its converse - proof and applications
• Solving real-life and exam problems using similarity and Pythagoras theorem
These notes cover every concept, theorem, and formula you need - complete with step-by-step solved examples, comparison tables, common mistakes to avoid, and categorised practice questions for board exam preparation.
A detailed and printable PDF version of these notes is attached below for your convenience. Download, save, and use it for your revision!
1. Introduction and Basic Definitions
Triangles are among the most fundamental shapes in geometry. In this chapter, we move beyond basic properties and study similarity - a concept that tells us when two triangles have the same shape (though not necessarily the same size). We also revisit and formally prove the Pythagoras theorem, one of the most celebrated results in all of mathematics.
1.1 Congruent vs Similar Figures
Two figures are congruent if they have exactly the same shape AND the same size. Two figures are similar if they have exactly the same shape but may differ in size. All congruent figures are similar, but not all similar figures are congruent.
Property | Congruent Figures | Similar Figures |
Shape | Same | Same |
Size | Same | May be different |
Corresponding angles | Equal | Equal |
Corresponding sides | Equal | Proportional |
Symbol | ≅ | ~ |
1.2 Similar Polygons
Two polygons of the same number of sides are similar if: (i) their corresponding angles are equal, and (ii) their corresponding sides are in the same ratio (i.e., proportional). Both conditions must hold simultaneously.
>> Key Insight For triangles, it turns out that we do NOT need to verify all six conditions. Just two or three conditions are sufficient to prove similarity. These special shortcuts are called the similarity criteria. |
2. Key Concepts and Theorems
2.1 Basic Proportionality Theorem (BPT) - Thales Theorem
This is the foundational theorem of this chapter. It relates a line drawn parallel to one side of a triangle with the other two sides.
>> Theorem 1 - Basic Proportionality Theorem If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then it divides the other two sides in the same ratio. In triangle ABC, if DE is parallel to BC and D is on AB and E is on AC, then: AD/DB = AE/EC |
>> Theorem 2 - Converse of BPT If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. If AD/DB = AE/EC, then DE is parallel to BC. |
2.2 Criteria for Similarity of Triangles
There are three main criteria used to establish that two triangles are similar. Unlike congruence, similarity requires only a subset of conditions to be verified.
Criterion | Full Name | Conditions Required |
AA Similarity | Angle-Angle | Two pairs of corresponding angles are equal |
SSS Similarity | Side-Side-Side | All three pairs of corresponding sides are proportional |
SAS Similarity | Side-Angle-Side | Two pairs of sides proportional AND included angle equal |
2.3 AA Similarity Criterion
>> Theorem 3 - AA Similarity If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. (The third angle is automatically equal since angle sum of a triangle is 180 degrees.) |
2.4 SSS Similarity Criterion
>> Theorem 4 - SSS Similarity If the corresponding sides of two triangles are proportional (in the same ratio), then the triangles are similar. If AB/DE = BC/EF = CA/FD, then triangle ABC ~ triangle DEF. |
2.5 SAS Similarity Criterion
>> Theorem 5 - SAS Similarity If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar. If angle A = angle D and AB/DE = AC/DF, then triangle ABC ~ triangle DEF. |
2.6 Areas of Similar Triangles
>> Theorem 6 - Area Ratio Theorem The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. If triangle ABC ~ triangle PQR, then: Area(ABC) / Area(PQR) = (AB/PQ)^2 = (BC/QR)^2 = (CA/RP)^2 |
2.7 Pythagoras Theorem
>> Theorem 7 - Pythagoras Theorem In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. If angle B = 90 degrees in triangle ABC, then: AC^2 = AB^2 + BC^2 |
>> Theorem 8 - Converse of Pythagoras Theorem If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. If AC^2 = AB^2 + BC^2, then angle B = 90 degrees. |
3. Core Formulas and Results
3.1 Basic Proportionality Theorem Formula
BPT (Thales Theorem) DE || BC => AD/DB = AE/EC
Equivalent forms: AD/AB = AE/AC (fraction of full side) DB/AB = EC/AC (remaining fraction) |
3.2 Similarity Ratio
Ratio of Corresponding Parts in Similar Triangles If triangle ABC ~ triangle DEF, then:
AB/DE = BC/EF = CA/FD = k (scale factor)
Perimeter(ABC) / Perimeter(DEF) = k Area(ABC) / Area(DEF) = k^2 |
3.3 Pythagoras Theorem
Pythagoras Theorem and Key Results Right angle at B: AC^2 = AB^2 + BC^2
Pythagorean Triplets (common ones): 3, 4, 5 (3^2 + 4^2 = 5^2) 5, 12, 13 (5^2 + 12^2 = 13^2) 8, 15, 17 (8^2 + 15^2 = 17^2) 7, 24, 25 (7^2 + 24^2 = 25^2) |
3.4 Altitude on Hypotenuse Result
In a right triangle, altitude from right angle to hypotenuse: If BD is altitude from B to AC in right triangle ABC (right angle at B):
BD^2 = AD x DC AB^2 = AD x AC BC^2 = DC x AC |
4. Solved Examples
4.1 BPT and Proportionality
Example 4.1 - Using BPT to find unknown lengths In triangle ABC, DE || BC. If AD = 3 cm, DB = 5 cm, AE = 4.5 cm, find EC.
By BPT (Thales Theorem): AD/DB = AE/EC 3/5 = 4.5/EC EC = (4.5 x 5) / 3 = 22.5 / 3 = 7.5 cm
Answer: EC = 7.5 cm |
Example 4.2 - Verifying if a line is parallel using Converse of BPT In triangle PQR, S is on PQ and T is on PR. PS = 3, SQ = 6, PT = 4, TR = 8. Check whether ST || QR.
PS/SQ = 3/6 = 1/2 PT/TR = 4/8 = 1/2
Since PS/SQ = PT/TR, by the Converse of BPT, ST is parallel to QR.
Answer: ST || QR |
4.2 Similarity of Triangles
Example 4.3 - Proving triangles similar using AA criterion In triangle ABC, angle A = 60 degrees. In triangle DEF, angle D = 60 degrees and angle E = angle B = 50 degrees. Prove triangle ABC ~ triangle DEF.
In triangle ABC: angle A = 60, angle B = 50, so angle C = 70 degrees In triangle DEF: angle D = 60, angle E = 50, so angle F = 70 degrees
Corresponding angles: angle A = angle D = 60, angle B = angle E = 50 By AA Similarity criterion: triangle ABC ~ triangle DEF. |
Example 4.4 - Finding sides using similarity ratio Triangle ABC ~ Triangle PQR. AB = 6 cm, BC = 8 cm, AC = 10 cm, PQ = 9 cm. Find QR and PR.
Scale factor k = PQ/AB = 9/6 = 3/2
QR = BC x k = 8 x 3/2 = 12 cm PR = AC x k = 10 x 3/2 = 15 cm
Answer: QR = 12 cm, PR = 15 cm |
4.3 Areas of Similar Triangles
Example 4.5 - Using the Area Ratio Theorem Two similar triangles have corresponding sides 6 cm and 9 cm. If the area of the smaller triangle is 36 sq cm, find the area of the larger triangle.
Area ratio = (corresponding side ratio)^2 Area(larger) / Area(smaller) = (9/6)^2 = (3/2)^2 = 9/4
Area(larger) = 36 x (9/4) = 81 sq cm
Answer: Area of larger triangle = 81 sq cm |
4.4 Pythagoras Theorem Applications
Example 4.6 - Applying Pythagoras Theorem A ladder 13 m long leans against a wall. Its foot is 5 m from the base of the wall. How high up the wall does the ladder reach?
Let the height be h. By Pythagoras: h^2 + 5^2 = 13^2 h^2 = 169 - 25 = 144 h = 12 m
Answer: The ladder reaches 12 m up the wall. |
Example 4.7 - Using Converse of Pythagoras Theorem Check if a triangle with sides 7 cm, 24 cm, and 25 cm is right-angled.
Check: 7^2 + 24^2 = 49 + 576 = 625 = 25^2 OK
Since the square of the longest side equals the sum of squares of the other two, by the Converse of Pythagoras Theorem, the triangle is right-angled.
Answer: Yes, it is a right-angled triangle (right angle opposite the 25 cm side). |
5. Applications and Special Cases
5.1 Real-Life Applications
• Surveying: Measuring inaccessible distances across rivers or valleys using similar triangle ratios.
• Architecture: Scaling blueprints up or down while maintaining proportional dimensions.
• Navigation: Calculating distances using angle measurements (triangulation).
• Construction: Using Pythagorean triplets to set right angles in building foundations.
• Art and Design: Using similar figures to create scaled representations and perspective drawings.
5.2 Important Special Cases
The following are special results that frequently appear in board exam questions:
• In a right triangle, the altitude from the right angle to the hypotenuse creates two triangles, each similar to the original and to each other.
• If two triangles are similar with scale factor k, all corresponding lengths (sides, altitudes, medians, angle bisectors) are in ratio k, but areas are in ratio k^2.
• A line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length (Mid-Point Theorem - a direct consequence of BPT).
• For equilateral triangles: all equilateral triangles are similar to each other.
• For squares: all squares are similar to each other. But rectangles and rhombuses are NOT always similar to each other.
5.3 Mid-Point Theorem as a Corollary of BPT
>> Mid-Point Theorem The line segment joining the midpoints of any two sides of a triangle is parallel to and half the length of the third side. If M and N are midpoints of AB and AC in triangle ABC, then MN || BC and MN = (1/2) BC. This is a direct consequence of the Basic Proportionality Theorem. |
6. Formula Summary Table
All key results for Triangles at a glance:
Result / Theorem | Statement | When to Use |
BPT | AD/DB = AE/EC if DE || BC | Line parallel to a side divides other sides proportionally |
Converse of BPT | DE || BC if AD/DB = AE/EC | Prove a line is parallel to a side |
AA Similarity | Two equal angles => similar triangles | Quickest similarity proof |
SSS Similarity | AB/DE = BC/EF = CA/FD | All sides proportional |
SAS Similarity | AB/DE = AC/DF and included angle equal | Two sides + included angle |
Area Ratio | Area ratio = (side ratio)^2 | Areas of similar triangles |
Pythagoras | AC^2 = AB^2 + BC^2 (right angle at B) | Right-angled triangles |
Converse of Pythagoras | AC^2 = AB^2 + BC^2 => angle B = 90 | Proving a right angle |
Altitude on hypotenuse | BD^2 = AD x DC | Right triangle altitude result |
Scale factor k | Sides in ratio k, areas in ratio k^2 | Comparing similar triangles |
7. Key Theorems and Properties
>> Theorem Summary - BPT Family 1. BPT: DE || BC => AD/DB = AE/EC 2. Converse: AD/DB = AE/EC => DE || BC 3. Corollary: If D and E are midpoints of AB and AC, then DE || BC and DE = (1/2)BC |
>> Theorem Summary - Similarity Criteria 1. AA: Two equal angles 2. SSS: Three pairs of sides proportional 3. SAS: Two sides proportional and included angle equal Note: AAA and SSS for congruence are DIFFERENT from similarity criteria - do not confuse them. |
>> Theorem Summary - Pythagoras Family 1. Pythagoras: Right angle at B => AC^2 = AB^2 + BC^2 2. Converse: AC^2 = AB^2 + BC^2 => angle B = 90 degrees 3. Altitude result: BD^2 = AD x DC where BD is altitude on hypotenuse AC |
7.1 Important Properties of Similar Triangles
Property | Ratio |
Corresponding sides | k (scale factor) |
Corresponding altitudes | k |
Corresponding medians | k |
Corresponding angle bisectors | k |
Perimeters | k |
Areas | k^2 |
8. Common Mistakes and Exam Tips
8.1 Frequent Errors to Avoid
1. Writing the similarity as triangle ABC ~ triangle DEF without ensuring the vertices are in the correct corresponding order.
2. Confusing the area ratio with the side ratio. Remember: area ratio = (side ratio)^2, NOT the side ratio itself.
3. Applying BPT without first confirming that the line is actually parallel to the base.
4. Using SAS similarity but picking the wrong angle (must be the INCLUDED angle between the two proportional sides).
5. Forgetting to take the square root when the question gives the area ratio and asks for the side ratio.
6. Stating Pythagoras without specifying which angle is the right angle.
8.2 Exam Strategy Tips
• Always label vertices carefully when writing similarity statements - order matters.
• In proof questions, state the theorem name clearly (e.g., 'By AA Similarity criterion').
• Learn the common Pythagorean triplets: (3,4,5), (5,12,13), (8,15,17), (7,24,25).
• For area ratio questions, identify the side ratio first, then square it.
• BPT proof and Pythagoras theorem proof are directly asked - memorise both proofs.
• Draw a neat, labelled diagram for every geometry question - it helps avoid errors and earns presentation marks.
9. Practice Questions
Questions are organised by marks as in CBSE board examinations. Attempt all categories for complete preparation.
9.1 - 1 Mark Questions (MCQ / Very Short Answer)
1. If triangle ABC ~ triangle PQR and AB/PQ = 3/4, find Area(ABC)/Area(PQR).
2. In triangle DEF, if GH || EF where G is on DE and H is on DF, and DG = 4 cm, GE = 6 cm, DH = 5 cm, find HF.
3. State the Basic Proportionality Theorem.
4. A triangle has sides 5 cm, 12 cm, and 13 cm. Is it a right triangle? Give reason.
5. If two similar triangles have a scale factor of 3:5, what is the ratio of their areas?
6. In triangle ABC, if DE || BC, AD = 2 cm, DB = 3 cm and AE = 3 cm. Find EC.
9.2 - 3 Mark Questions (Short Answer)
1. In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If AD/DB = 3/5 and AE = 4.5 cm, find AC.
2. Prove that if two triangles are similar, then the ratio of their areas equals the square of the ratio of their corresponding sides.
3. In the given figure, triangle OAB ~ triangle OCD. If OA = 6 cm, OC = 4 cm and AB = 9 cm. Find CD.
4. Two poles of height 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
5. In triangle ABC, the altitudes BE and CF are equal. Prove that triangle ABE ~ triangle ACF and hence prove that triangle ABC is isosceles.
9.3 - 5 Mark Questions (Long Answer)
1. Prove the Basic Proportionality Theorem (Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Also prove its converse.
2. Prove the Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
3. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, prove that the angle opposite the first side is a right angle (Converse of Pythagoras Theorem).
4. Through the mid-point M of side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2 BL.
5. In a right triangle ABC, right-angled at B, if BD is an altitude to the hypotenuse AC, prove that (i) triangle ABD ~ triangle ABC, (ii) triangle BDC ~ triangle ABC, (iii) BD^2 = AD x DC.
CBSE Class 10 Syllabus |
CBSE Class 10 Notes |

