Class 10 Maths Chapter 6: Triangles — Important Questions & Sample Paper
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Reviewed by qpaper's CBSE curriculum team · Edited by Mohit · Updated June 2026
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Chapter 6 of CBSE Class 10 Mathematics, ‘Triangles’, deepens students’ understanding of geometric figures by exploring the concept of similarity. This chapter builds on earlier knowledge of congruency and introduces conditions under which two triangles are similar but not necessarily congruent. Key learning points include the criteria for similarity: AAA (Angle-Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side), alongside the foundational Basic Proportionality Theorem (Thales theorem) and its converse. Students learn that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally, a principle that appears in many exam questions.
The chapter also covers the relationship between the areas of similar triangles—the ratio of their areas equals the square of the ratio of corresponding sides—and extends similarity to practical problems involving heights and distances. Right-angled triangles receive special attention with the Pythagoras theorem and its converse, which are crucial for solving numeric and proof-based problems. Common exam questions ask students to apply criteria to prove similarity, find unknown sides using proportions, calculate area ratios, or solve geometric construction-based problems. For instance, a typical problem might give two triangles with proportional sides and an included angle, expecting students to recognize the SAS similarity criterion. Understanding these concepts thoroughly equips students to tackle both direct and application-level questions in the CBSE board exams.
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Mathematics — Triangles
SECTION A
- 1.1
In triangle ABC, right-angled at B, BD is drawn perpendicular to AC. If AB = 6 cm and BC = 8 cm, then AD : DC equals:
(a) 9:16(b) 16:9(c) 3:4(d) 4:3 - 2.1
In triangle ABC, DE is drawn parallel to BC, intersecting AB at D and AC at E. If AD = 2 cm, DB = 3 cm, and AE = 3 cm, then EC equals:
(a) 4.5 cm(b) 5 cm(c) 6 cm(d) 4 cm - 3.1
In triangle ABC, D is a point on AB and E on AC such that DE is parallel to BC. If AD:DB = 1:2 and the area of triangle ADE is 10 cm², then the area of quadrilateral BDEC is:
(a) 20 cm²(b) 30 cm²(c) 60 cm²(d) 80 cm²
+ 41 more questions in the full paper
Generate full paperMarks distribution & blueprint
In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:
| Question type | Marks each | In our bank |
|---|---|---|
| Multiple Choice (MCQ) | 1 mark | 13 |
| Assertion–Reason | 1 mark | 6 |
| Short Answer | 2 marks | 8 |
| Short Answer | 3 marks | 6 |
| Long Answer | 5 marks | 5 |
| Case Study | 4 marks | 6 |
44 original, exam-style questions in our bank for this chapter — with answers.
Important & sample questions (with answers)
Real, exam-style questions to practise and revise — each with its answer. Generate a full paper for unlimited more.
- Multiple Choice (MCQ)
Q1. In triangle ABC, right-angled at B, BD is drawn perpendicular to AC. If AB = 6 cm and BC = 8 cm, then AD : DC equals:
1 mark(A) 9:16(B) 16:9(C) 3:4(D) 4:3▸ Answer▾ Answer
9:16
- Multiple Choice (MCQ)
Q2. In triangle ABC, DE is drawn parallel to BC, intersecting AB at D and AC at E. If AD = 2 cm, DB = 3 cm, and AE = 3 cm, then EC equals:
1 mark(A) 4.5 cm(B) 5 cm(C) 6 cm(D) 4 cm▸ Answer▾ Answer
4.5 cm
- Multiple Choice (MCQ)
Q3. In triangle ABC, D is a point on AB and E on AC such that DE is parallel to BC. If AD:DB = 1:2 and the area of triangle ADE is 10 cm², then the area of quadrilateral BDEC is:
1 mark(A) 20 cm²(B) 30 cm²(C) 60 cm²(D) 80 cm²▸ Answer▾ Answer
80 cm²
- Multiple Choice (MCQ)
Q4. In triangle ABC, AB = 4 cm, AC = 6 cm, and ∠A = 50°. In triangle PQR, PQ = 8 cm, PR = 12 cm, and ∠P = 50°. The triangles are similar by which criterion?
1 mark(A) SSS similarity(B) SAS similarity(C) AAA similarity(D) They are not similar▸ Answer▾ Answer
SAS similarity
- Assertion–Reason
Q5. Assertion (A): In triangle ABC, D and E are points on AB and AC such that AD/DB = AE/EC, then DE || BC. Reason (R): The Basic Proportionality Theorem states that if a line is parallel to one side of a triangle, it divides the other two sides proportionally.
1 mark(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.▸ Answer▾ Answer
Both A and R are true but R is not the correct explanation of A.
- Short Answer
Q6. Two similar triangles have their areas in the ratio 9 : 16. Find the ratio of their corresponding sides.
2 marks▸ Answer▾ Answer
3 : 4
- Short Answer
Q7. In an equilateral triangle ABC, D is a point on BC such that BD = 1/3 BC. Prove that 9 AD² = 7 AB².
2 marks▸ Answer▾ Answer
9 AD² = 7 AB²
- Short Answer
Q8. The perimeters of two similar triangles are 30 cm and 45 cm respectively. If the area of the smaller triangle is 40 cm², find the area of the larger triangle.
3 marks▸ Answer▾ Answer
90 cm²
- Short Answer
Q9. In triangle XYZ, P and Q are points on XY and XZ respectively such that PQ is parallel to YZ. If XP = 2x, PY = 3x − 1, XQ = x + 2, and QZ = 4x − 3, find the value of x and the length of XY.
3 marks▸ Answer▾ Answer
x = 2, XY = 9 units
- Long Answer
Q10. Two vertical poles of heights a meters and b meters stand on a level ground with their bases p meters apart. The line segment joining the top of the first pole to the base of the second and the line segment joining the top of the second pole to the base of the first intersect at a point. Find the height of this intersection point from the ground in terms of a and b. Prove your result.
5 marks▸ Answer▾ Answer
The height is ab/(a+b) meters.
- Long Answer
Q11. If two triangles are similar, prove that the ratio of their areas is equal to the square of the ratio of their corresponding sides.
5 marks▸ Answer▾ Answer
If ΔABC ~ ΔPQR with similarity ratio k, then area(ΔABC) / area(ΔPQR) = k².
- Case Study
Q12. Ravi uses a mirror on the ground to measure the height of a tree. He places the mirror at point M. The tree is at point T, and Ravi stands at point R such that his eyes at point E see the top of the tree in the mirror. TM = 6 m, MR = 0.75 m, and the height of Ravi's eyes from the ground (ER) is 1.5 m.
4 marks- (i) Show that the two triangles formed are similar and state the similarity criterion used.2 marks
- (ii) Find the height of the tree.2 marks
▸ Answer▾ Answer
The triangles are similar by AA criterion. Height of tree = 12 m.
Frequently asked questions
What is the Basic Proportionality Theorem and how is it tested in exams?
The Basic Proportionality Theorem (Thales theorem) states that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. Its converse holds as well. Exam questions often ask to find an unknown length given a parallel line segment, or to prove that a certain configuration yields proportional sides.
Which criteria are used to prove triangle similarity in CBSE Class 10?
The three criteria are AAA (angle-angle-angle, often reduced to AA as the third angle automatically matches), SSS (side-side-side, all three sides in proportion), and SAS (side-angle-side, two sides in proportion and the included angle equal). Problems may ask to identify the correct criterion or prove similarity using given measurements.
How do you solve problems involving areas of similar triangles?
For similar triangles, the ratio of their areas equals the square of the ratio of any pair of corresponding sides. If the side ratio is m:n, area ratio is m²:n². This is applied to compute unknown areas, side lengths from given areas, or to prove certain geometric relationships.
What is the importance of Pythagoras theorem in this chapter?
Pythagoras theorem is a special property of right-angled triangles that appears frequently in the CBSE exam. It states that the square of the hypotenuse equals the sum of the squares of the other two sides. The converse is also used to prove whether a triangle is right-angled. Questions often involve finding a missing side, applying the theorem in composite figures, or proving the result in a given context.
More chapters
- Ch 1: Real Numbers
- Ch 2: Polynomials
- Ch 3: Pair of Linear Equations in Two Variables
- Ch 4: Quadratic Equations
- Ch 5: Arithmetic Progressions
- Ch 6: Triangles
- Ch 7: Coordinate Geometry
- Ch 8: Introduction to Trigonometry
- Ch 9: Some Applications of Trigonometry
- Ch 10: Circles
- Ch 11: Areas Related to Circles
- Ch 12: Surface Areas and Volumes
- Ch 13: Statistics
- Ch 14: Probability