NCERT Solutions Class 10 Maths Chapter 6 helps you master Triangles through step-by-step solutions covering similarity criteria (AAA, SAS, SSA), Basic Proportionality Theorem, Pythagoras Theorem, and area relationships. You’ll learn how to prove triangles similar, calculate unknown sides using proportionality, and apply these concepts to solve real-world problems involving heights, distances, and geometric constructions that frequently appear in CBSE board exams.
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All exercises with step-by-step solutions | Updated 2025-26 | Free Download
Download PDF (Free)NCERT Solutions Class 10 Maths Chapter 6 Triangles – Complete Guide
NCERT Class 10 Chapter 6 – Triangles is a fundamental chapter that builds upon your understanding of geometry from Class 9. You will dive deep into the concept of similarity of triangles, learning how triangles can have the same shape but different sizes. This chapter carries 5 marks weightage in CBSE board exams and typically includes 2-3 questions ranging from MCQs to long answer problems worth 3-4 marks each.
📊 CBSE Class 10 Maths Chapter 6 – Exam Weightage & Marking Scheme
| CBSE Board Marks | 5 Marks |
| Unit Name | Geometry |
| Difficulty Level | Medium |
| Importance | Medium |
| Exam Types | CBSE Board, State Boards |
| Typical Questions | 1-2 questions |
You’ll explore the Basic Proportionality Theorem (Thales’ Theorem) and its converse, which form the foundation for understanding how parallel lines divide the sides of triangles proportionally. The chapter covers four crucial criteria for triangle similarity: AAA, AA, SSS, and SAS similarity criteria. You’ll learn to identify and prove when two triangles are similar, a skill that appears frequently in both board exams and competitive tests.
The chapter also includes the areas of similar triangles theorem, which states that the ratio of areas equals the square of the ratio of corresponding sides. You’ll revisit and prove the Pythagoras Theorem using similarity concepts, along with its converse. These theorems have practical applications in architecture, engineering, navigation, and map reading, making this chapter highly relevant beyond academics.
Quick Facts – Class 10 Chapter 6
| 📖 Chapter Number | Chapter 6 |
| 📚 Chapter Name | Triangles |
| ✏️ Total Exercises | 3 Exercises |
| ❓ Total Questions | 29 Questions |
| 📅 Updated For | CBSE Session 2025-26 |
Mastering Triangles is essential as it connects to coordinate geometry, trigonometry, and mensuration in later chapters. The problem-solving techniques you develop here will strengthen your logical reasoning and proof-writing abilities. With consistent practice of NCERT solutions and previous year CBSE questions, you’ll confidently tackle any triangle-related problem in your board exams and build a strong foundation for higher mathematics.
NCERT Solutions Class 10 Maths Chapter 6 – All Exercises PDF Download
Download exercise-wise NCERT Solutions PDFs for offline study
| Exercise No. | Topics Covered | Download PDF |
|---|---|---|
| Exercise 6.1 | Complete step-by-step solutions for 3 questions | 📥 Download PDF |
| Exercise 6.2 | Complete step-by-step solutions for 10 questions | 📥 Download PDF |
| Exercise 6.3 | Complete step-by-step solutions for 16 questions | 📥 Download PDF |
Triangles – Key Formulas & Concepts
Quick reference for CBSE exams
| Formula | Description | When to Use |
|---|---|---|
| Basic Proportionality Theorem (BPT) / Thales’ Theorem If \(DE \parallel BC\), then \(\frac{AD}{DB} = \frac{AE}{EC}\) | If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. Note: Make sure you are using corresponding segments (AD with DB, and AE with EC). The converse is also true. | When a line is parallel to a side of a triangle and you need to find the ratio of segments on the other two sides. |
| Converse of BPT If \(\frac{AD}{DB} = \frac{AE}{EC}\), then \(DE \parallel BC\) | If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Note: Important for proving lines are parallel. Check if the ratios are EXACTLY equal before concluding parallelism. | To prove that a line is parallel to one side of a triangle when the ratios of the segments on the other two sides are equal. |
| AAA Similarity Criterion If \(\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R\), then \(\triangle ABC \sim \triangle PQR\) | If all three angles of one triangle are equal to the corresponding three angles of another triangle, then the two triangles are similar. Note: AA similarity is sufficient (third angle is automatically equal if two angles are equal). Order of vertices matters! | To prove triangles are similar when you know (or can prove) that all three pairs of corresponding angles are equal. |
| SAS Similarity Criterion If \(\angle A = \angle P\) and \(\frac{AB}{PQ} = \frac{AC}{PR}\), then \(\triangle ABC \sim \triangle PQR\) | 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. Note: The angle must be *included* between the two sides that are in proportion. | To prove triangles are similar when you know one angle is equal and the sides including that angle are in proportion. |
| SSS Similarity Criterion If \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}\), then \(\triangle ABC \sim \triangle PQR\) | If three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar. Note: Check all three ratios. Remember to match corresponding sides correctly. | To prove triangles are similar when you know that all three pairs of corresponding sides are in proportion. |
| Area Ratio Theorem If \(\triangle ABC \sim \triangle PQR\), then \(\frac{Area(\triangle ABC)}{Area(\triangle PQR)} = (\frac{AB}{PQ})^2 = (\frac{BC}{QR})^2 = (\frac{CA}{RP})^2\) | The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Note: Don’t forget to SQUARE the ratio of the sides. Can be used to find unknown areas or side lengths. | When you know the triangles are similar and you need to find the ratio of their areas, or vice-versa. |
| Pythagoras Theorem In a right-angled triangle, \(AC^2 = AB^2 + BC^2\) | In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Note: Hypotenuse (AC) is always opposite the right angle. Make sure you identify it correctly. | To find the length of one side of a right-angled triangle when you know the lengths of the other two sides. |
| Converse of Pythagoras Theorem If \(AC^2 = AB^2 + BC^2\), then \(\angle B = 90^\circ\) | If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Note: Check if the longest side squared equals the sum of the squares of the other two sides. This is the only way to prove it’s a right triangle if you don’t know any angles. | To prove that a triangle is a right-angled triangle when you know the lengths of all three sides. |
| Area of an Equilateral Triangle \(Area = \frac{\sqrt{3}}{4}a^2\) | Calculate area of an equilateral triangle, where a is side length Note: Only for equilateral triangles | Direct calculation of equilateral triangle area given side length |
| Ratio of Medians & Sides in Similar Triangles If \(\triangle ABC \sim \triangle PQR\) and AD and PS are medians, then \(\frac{AD}{PS} = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}\) | The ratio of corresponding medians in similar triangles is equal to the ratio of corresponding sides. Note: Applies to altitudes, angle bisectors as well. | Problems involving similar triangles and their medians. |
Frequently Asked Questions – NCERT Class 10 Maths Chapter 6
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