Class 10 Maths Chapter 6 Triangles Notes With Free PDF Download

Class 10 Maths Chapter 6 Triangles Notes Free PDF Download is prepared based on the latest CBSE and NCERT syllabus. These notes will help in school exams, board exams and quick revision. They help students to understand the chapter clearly, revise faster and prepare for exams with confidence.

Table of Contents

• What Are Similar Figures?
• Similar Triangles
• Basic Proportionality Theorem (BPT)
• Criteria for Similarity of Triangles
• Areas of Similar Triangles
• Pythagoras Theorem
• Solved Examples on Class 10 Maths Chapter 6 Triangles Notes
• Practice Questions on Class 10 Maths Chapter 6 Triangles Notes
• Important Table of Formulas
• Previous Years Important Questions on Class 10 Maths Chapter 6 Triangles Notes

What Are Similar Figures?

Two figures are called similar if they have the same shape but not necessarily the same size.

• All circles are similar to each other
  
• All squares are similar to each other
  
• Equilateral triangles are always similar to each other
  

Similar vs Congruent

Example:

Two triangles ABC and DEF where:

Congruent: AB = DE, BC = EF, AC = DF (sides equal)

Similar: AB/DE = BC/EF = AC/DF (sides proportional)

Key Property of Similar Figures

When two figures are similar:

1. Corresponding angles are equal

2. Corresponding sides are proportional (in the same ratio)

If △ABC ~ △DEF (Triangle ABC similar to Triangle DEF)

Then:

∠A = ∠D

∠B = ∠E

∠C = ∠F

And:

AB/DE = BC/EF = AC/DF = k (constant ratio)

k is called the scale factor

Similar Triangles

Two triangles are similar if:

• Their corresponding angles are equal

• Their corresponding sides are in the same ratio

Notation:

△ABC ~ △DEF means "Triangle ABC is similar to Triangle DEF"

Important: The order of letters matters! When writing ABC ~ DEF:

A corresponds to D

B corresponds to E

C corresponds to F

Understanding Scale Factor

The scale factor tells you how many times bigger or smaller one triangle is compared to the other.

If AB/DE = BC/EF = AC/DF = 2

This means Triangle ABC is 2 times bigger than Triangle DEF

Example:

△ABC ~ △PQR

AB = 6 cm, PQ = 3 cm

BC = 8 cm, QR = 4 cm

AC = 10 cm, PR = 5 cm

Ratio = 6/3 = 8/4 = 10/5 = 2

Scale factor = 2

Properties of Similar Triangles

Property 1: If △ABC ~ △DEF, then all three ratios are equal

AB/DE = BC/EF = CA/FD

Property 2: Similarity is reflexive

Every triangle is similar to itself

△ABC ~ △ABC

Property 3: Similarity is symmetric

If △ABC ~ △DEF, then △DEF ~ △ABC

Property 4: Similarity is transitive

If △ABC ~ △DEF and △DEF ~ △PQR

Then △ABC ~ △PQR

Basic Proportionality Theorem (BPT)

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.

If in △ABC, DE || BC where D is on AB and E is on AC:

AD/DB = AE/EC

DE || BC

Therefore: AD/DB = AE/EC

Proof of BPT

Given:

• △ABC with DE || BC

• D on AB, E on AC

To prove: AD/DB = AE/EC

Construction: Draw DM ⊥ AC and EN ⊥ AB. Join BE and CD.

Proof:

Area of △ADE = (1/2) × AD × EN  ...(1)

Area of △BDE = (1/2) × DB × EN  ...(2)

Dividing (1) by (2):

Area(△ADE)/Area(△BDE) = AD/DB  ...(3)

Similarly:

Area of △ADE = (1/2) × AE × DM  ...(4)

Area of △CDE = (1/2) × EC × DM  ...(5)

Area(△ADE)/Area(△CDE) = AE/EC  ...(6)

Since △BDE and △CDE are on the same base DE

and between the same parallels DE and BC:

Area(△BDE) = Area(△CDE)  ...(7)

From (3), (6), and (7):

AD/DB = AE/EC  (Proved!)

Converse of BPT

Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Mathematical form:

If in △ABC:

AD/DB = AE/EC

Then DE || BC

Important Results from BPT

When DE || BC in △ABC:

Result 1: AD/DB = AE/EC

Result 2: AB/AD = AC/AE

(taking the whole side in numerator)

Result 3: DB/AB = EC/AC

Result 4: AD/AB = AE/AC

Solved Example on BPT

Example: In △ABC, DE || BC. AD = 4 cm, DB = 6 cm, AE = 3 cm. Find EC.

Solution:

By BPT:

AD/DB = AE/EC

4/6 = 3/EC

EC = (3 × 6)/4

EC = 18/4

EC = 4.5 cm

Answer: EC = 4.5 cm

Criteria for Similarity of Triangles

There are three important criteria to prove two triangles are similar. You don't need to check all six parts (3 sides + 3 angles). These shortcuts save time!

1. AA Similarity Criterion (Angle-Angle)

Theorem: If two angles of one triangle are equal to two angles of another triangle, then the two triangles are similar.

Condition:

If ∠A = ∠D and ∠B = ∠E

Then △ABC ~ △DEF

Why only two angles?

Because if two angles match, the third automatically matches too. (All angles of a triangle sum to 180°)

Example:

In △ABC and △DEF:

∠A = ∠D = 60°

∠B = ∠E = 70°

Therefore: ∠C = ∠F = 50° (automatically)

So △ABC ~ △DEF (by AA)

  70° 50°         70° 50°

△ABC ~ △DEF

2. SSS Similarity Criterion (Side-Side-Side)

Theorem: If the corresponding sides of two triangles are in the same ratio, then the triangles are similar.

Condition:

If AB/DE = BC/EF = CA/FD

Then △ABC ~ △DEF

Example:

In △ABC: AB = 4, BC = 6, CA = 8

In △DEF: DE = 2, EF = 3, FD = 4

Check:

AB/DE = 4/2 = 2

BC/EF = 6/3 = 2

CA/FD = 8/4 = 2

All ratios equal! → △ABC ~ △DEF (by SSS)

3. SAS Similarity Criterion (Side-Angle-Side)

Theorem: If one angle of a triangle equals one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.

Condition:

If AB/DE = AC/DF and ∠A = ∠D

Then △ABC ~ △DEF

Important: The equal angle must be the angle BETWEEN the proportional sides!

Example:

In △ABC and △DEF:

AB = 4, DE = 2 (ratio = 2)

AC = 6, DF = 3 (ratio = 2)

∠A = ∠D = 50°

The angle is between the proportional sides

→ △ABC ~ △DEF (by SAS)

Comparison of Criteria

Areas of Similar Triangles

The ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides.

Mathematical form:

If △ABC ~ △DEF:

Area(△ABC)/Area(△DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)²

If sides are in ratio 2:1, then areas are in ratio 4:1 (square of 2:1)

If sides are in ratio 3:2, then areas are in ratio 9:4

Proof of the Theorem

Given: △ABC ~ △DEF

To prove: Area(△ABC)/Area(△DEF) = (BC/EF)²

Construction: Draw AM ⊥ BC and DN ⊥ EF

Proof:

Area(△ABC) = (1/2) × BC × AM  ...(1)

Area(△DEF) = (1/2) × EF × DN  ...(2)

 Dividing:

Area(△ABC)/Area(△DEF) = (BC × AM)/(EF × DN)  ...(3)

 Since △ABC ~ △DEF:

∠B = ∠E

 In △ABM and △DEN:

∠AMB = ∠DNE = 90° (construction)

∠B = ∠E (given)

 So △ABM ~ △DEN (AA)

Therefore: AM/DN = AB/DE = BC/EF  ...(4)

Substituting (4) in (3):

Area(△ABC)/Area(△DEF) = (BC/EF) × (BC/EF) = (BC/EF)²

Results:

If △ABC ~ △DEF and AB/DE = k

Then:

Area(△ABC)/Area(△DEF) = k²

 Perimeter(△ABC)/Perimeter(△DEF) = k

(Perimeters are in the same ratio as sides, not squared!)

Example on Areas

Example: △ABC ~ △DEF. If AB = 6 cm and DE = 4 cm, find the ratio of their areas.

Solution:

AB/DE = 6/4 = 3/2

Area ratio = (AB/DE)² = (3/2)² = 9/4

Area(△ABC) : Area(△DEF) = 9 : 4

Answer: Areas are in ratio 9:4

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.

If ∠B = 90° in △ABC:

AC² = AB² + BC²

(Hypotenuse)² = (Base)² + (Height)²

AC² = AB² + BC²

Proof of Pythagoras Theorem

Given: △ABC, right-angled at B (∠B = 90°)

To prove: AC² = AB² + BC²

Construction: Draw BD ⊥ AC

Proof:

In △ADB and △ABC:

∠ADB = ∠ABC = 90°

∠A = ∠A (common)

By AA: △ADB ~ △ABC

Therefore: AD/AB = AB/AC

AB² = AD × AC  ...(1)

In △BDC and △ABC:

∠BDC = ∠ABC = 90°

∠C = ∠C (common)

By AA: △BDC ~ △ABC

Therefore: DC/BC = BC/AC

BC² = DC × AC  ...(2)

Adding (1) and (2):

AB² + BC² = AD × AC + DC × AC

AB² + BC² = AC(AD + DC)

AB² + BC² = AC × AC

AB² + BC² = AC²

(Proved!)

Converse of Pythagoras Theorem

Theorem: If in a triangle, the square of one side equals the sum of squares of the other two sides, then the angle opposite to the first side is a right angle.

Mathematical form:

If AC² = AB² + BC²

Then ∠B = 90°

Pythagorean Triplets

These are sets of three numbers that satisfy Pythagoras theorem.

Generating Pythagorean triplets:

For any number m > 1:

Triplet: (2m, m² - 1, m² + 1)

Example: m = 3

2m = 6, m² - 1 = 8, m² + 1 = 10

Check: 6² + 8² = 36 + 64 = 100 = 10²

Solved Examples on Class 10 Maths Chapter 6 Triangles Notes

Example 1: BPT Application

Question: In the figure, DE || BC. AD = 3 cm, DB = 5 cm, AE = 2.4 cm. Find EC.

Solution:

By BPT:

AD/DB = AE/EC

3/5 = 2.4/EC

EC = (2.4 × 5)/3

EC = 12/3

EC = 4 cm

Answer: EC = 4 cm

Example 2: Check for Parallel Lines

Question: In △ABC, D and E are points on AB and AC. AD = 2, DB = 4, AE = 3, EC = 6. Is DE || BC?

Solution:

AD/DB = 2/4 = 1/2

AE/EC = 3/6 = 1/2

Since AD/DB = AE/EC = 1/2

By Converse of BPT: DE || BC

Answer: Yes, DE || BC

Example 3: AA Similarity

Question: In △ABC and △DEF, ∠A = 50°, ∠B = 70°, ∠D = 50°, ∠E = 70°. Are the triangles similar? If yes, write the similarity.

Solution:

In △ABC:

∠A = 50°, ∠B = 70°

∠C = 180° - 50° - 70° = 60°

In △DEF:

∠D = 50°, ∠E = 70°

∠F = 180° - 50° - 70° = 60°

∠A = ∠D = 50°

∠B = ∠E = 70°

By AA Criterion: △ABC ~ △DEF

Answer: Yes, △ABC ~ △DEF (by AA)

Example 4: SSS Similarity

Question: In △ABC, AB = 6 cm, BC = 9 cm, CA = 12 cm. In △DEF, DE = 4 cm, EF = 6 cm, FD = 8 cm. Are the triangles similar?

Solution:

AB/DE = 6/4 = 3/2

BC/EF = 9/6 = 3/2

CA/FD = 12/8 = 3/2

AB/DE = BC/EF = CA/FD = 3/2

By SSS Criterion: △ABC ~ △DEF

Answer: Yes, △ABC ~ △DEF (by SSS, ratio = 3/2)

Example 5: Area of Similar Triangles

Question: △ABC ~ △PQR. BC = 8 cm and QR = 6 cm. If area of △PQR = 36 cm², find area of △ABC.

Solution:

Area(△ABC)/Area(△PQR) = (BC/QR)²

Area(△ABC)/36 = (8/6)²

Area(△ABC)/36 = 64/36

Area(△ABC) = 36 × 64/36

Area(△ABC) = 64 cm²

Answer: Area of △ABC = 64 cm²

Example 6: Pythagoras Theorem

Question: In △ABC, ∠B = 90°. AB = 7 cm, BC = 24 cm. Find AC.

Solution:

By Pythagoras Theorem:

AC² = AB² + BC²

AC² = 7² + 24²

AC² = 49 + 576

AC² = 625

AC = √625

AC = 25 cm

Answer: AC = 25 cm

Example 7: Converse of Pythagoras

Question: A triangle has sides 11 cm, 60 cm, and 61 cm. Is it a right-angled triangle?

Solution:

Check: (Smallest sides)² + (Second side)²

= 11² + 60²

= 121 + 3600

= 3721

 (Largest side)² = 61² = 3721

 Since 11² + 60² = 61²

 By Converse of Pythagoras:

This is a right angled triangle!

 Answer: Yes, it's a right angled triangle (right angle opposite to side 61 cm)

Example 8: Combined Application

Question: In △ABC, ∠ABC = 90° and BD ⊥ AC. Prove that BD² = AD × DC.

Solution:

In △ABD and △BCD:

∠ADB = ∠BDC = 90° (BD ⊥ AC)

 In △ABD and △ABC:

∠ADB = ∠ABC = 90°

∠A is common

 △ABD ~ △ABC (by AA)

Therefore: AD/BD = BD/BC ... (not what we need)

 Better approach:

In △BDA and △CDB:

∠BDA = ∠CDB = 90°

∠ABD + ∠DBC = 90° (since ∠ABC = 90°)

∠ABD + ∠A = 90° (since ∠ADB = 90°)

Therefore ∠DBC = ∠A

 So △BDA ~ △CDB (by AA)

 Therefore: BD/CD = AD/BD

 BD² = AD × DC (Proved!)

Example 9: Finding Missing Side Using Similarity

Question: In the given figure, △ACB ~ △APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm. Find AQ.

Solution:

△ACB ~ △APQ

Corresponding sides:

AC ↔ AP

CB ↔ PQ

BA ↔ QA

BC/PQ = BA/AQ

8/4 = 6.5/AQ

2 = 6.5/AQ

AQ = 6.5/2

AQ = 3.25 cm

Answer: AQ = 3.25 cm

Example 10: Three Stage Problem

Question: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to D such that DM = CM. Point D is joined to point B. Show that: a) △AMC ≅ △BMD b) ∠DBC is a right angle c) △DBC ≅ △ACB d) CM = (1/2)AB

Solution:

Part a) △AMC ≅ △BMD:

In △AMC and △BMD:

AM = BM (M is mid-point of AB)

CM = DM (given)

∠AMC = ∠BMD (vertically opposite angles)

By SAS: △AMC ≅ △BMD

Part b) ∠DBC is right angle:

From part a): △AMC ≅ △BMD

Therefore: ∠ACM = ∠BDM (CPCT)

These are alternate angles for DB and AC with BC as transversal

Therefore: DB || AC

∠DBC + ∠ACB = 180° (co-interior angles)

∠DBC + 90° = 180°

∠DBC = 90°

Part c) △DBC ≅ △ACB:

BC = BC (common)

DB = AC (from part a, CPCT)

∠DBC = ∠ACB = 90°

By SAS: △DBC ≅ △ACB

Part d) CM = (1/2)AB:

From part c): DC = AB (CPCT)

But CM = DM = (1/2)DC = (1/2)AB

Therefore: CM = (1/2)AB (Proved)

Practice Questions on Class 10 Maths Chapter 6 Triangles Notes

Section 1: Basic Concepts

Question 1: State the Basic Proportionality Theorem.

Question 2: In △ABC, D and E are points on AB and AC. If AD/DB = AE/EC, what can you conclude?

Question 3: Two triangles are similar with sides in ratio 3:5. What is the ratio of their areas?

Question 4: State the AA similarity criterion.

Question 5: Find the Pythagorean triplet where one member is 20.

Section 2: BPT Problems

Question 6: In △ABC, DE || BC. If AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm, find AE and EC.

Question 7: In △PQR, ST || QR. If PS = 3, SQ = 5, PT = 4.5 cm, find TR.

Question 8: In △ABC, DE || BC. If AD/AB = 2/5, find DE/BC.

Question 9: E and F are points on sides AB and AC of △ABC. EF || BC. If AE = 3.6 cm, EB = 2.4 cm, AC = 9 cm, find AF.

Question 10: In △ABC, DE || BC. BD = 9 cm, AD = 6 cm, AE = 8 cm. Find the length of EC.

Section 3: Similarity

Question 11: Prove that if a line divides two sides of a triangle proportionally, it is parallel to the third side.

Question 12: In △ABC and △DEF, AB/DE = BC/EF = CA/FD = 4/3. Are triangles similar? State the criterion.

Question 13: In right △ABC, ∠B = 90°. BD ⊥ AC. If AD = 9 and DC = 4, find BD.

Question 14: △ABC ~ △DEF. If AB = 12 cm, DE = 9 cm, find the ratio of their perimeters.

Question 15: The areas of two similar triangles are 64 cm² and 100 cm². If a side of the smaller triangle is 8 cm, find the corresponding side of the larger triangle.

Section 4: Pythagoras Theorem

Question 16: In △ABC, ∠C = 90°. AC = 5 cm, BC = 12 cm. Find AB.

Question 17: Check if triangle with sides 6, 8, 10 is right angled.

Question 18: A ladder 13 m long leans against a wall. If the foot of the ladder is 5 m from the wall, how high up the wall does it reach?

Question 19: In an equilateral triangle with side 2a, find the length of each altitude.

Question 20: In △ABC, AB = 9 cm, BC = 12 cm, AC = 15 cm. Is ∠B = 90°? Find the area.

Important Table of Formulas:

Similarity Criteria:

Important Ratios:

BPT: AD/DB = AE/EC (when DE || BC)

Area ratio: (side₁/side₂)²

Perimeter ratio: side₁/side₂

Scale factor: k = any corresponding side ratio

Pythagoras Theorem:

In right △ABC (∠B = 90°):

AC² = AB² + BC²

Hypotenuse² = Base² + Height²

Previous Years Important Questions on Class 10 Maths Chapter 6 Triangles Notes

1 Mark Questions:

• State Basic Proportionality Theorem

• Write the converse of Pythagoras theorem

• If △ABC ~ △DEF and AB/DE = 3/4, find area ratio

2 Mark Questions:

• In △ABC, DE || BC. Find missing sides using BPT

• Check if given triangle is right-angled using converse of Pythagoras

• Find the third side of right triangle using Pythagoras theorem

3 Mark Questions:

• Prove BPT

• Two similar triangles find sides or areas

• Applications of Pythagoras theorem in real problems

4 Mark Questions:

• Prove Pythagoras theorem using similarity

• Complex problems combining BPT and similarity

• Multi-step problems involving all concepts of chapter

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