Class 9 Mathematics Chapter 6: Triangles — Previous Year Questions (2020–2025) with Solutions

Reading Chapter 6 from your textbook is essential, but it's passive. When you solve past paper questions, you: 1. **Identify What Examiners Actually Test** — Board papers focus heavily on applying theorems (BPT, similarity criteria) to solve numerical problems, not just stating definitions. You'll notice that 3-mark questions almost always demand a two-step proof or construction argument. 2. **Train Your Proof-Writing Style** — Triangles questions require formal geometric proofs. By working PYQ, you learn the exact logical flow and notation examiners expect (e.g., 'In △ABC and △DEF, ∠A = ∠D (given), AB/DE = AC/DF (given), ∴ △ABC ~ △DEF by SAS similarity'). 3. **Spot Theorem Combinations** — Examiners often layer concepts: they might ask you to use BPT to find a side, then use that to prove similarity, then find the ratio of areas. Past papers show these combinations first. 4. **Manage Exam Anxiety** — When you've solved a similar question before, confidence rises. You spend less time panicking and more time executing. 5. **Catch Calculation Traps** — Many students correctly identify that two triangles are similar but then miscalculate the area ratio as 2:3 instead of 4:9 (not squaring the linear ratio). Repeated practice prevents this. The strategy: solve 2–3 previous year questions daily for one week, review where you went wrong, then move to new unseen problems. This cycle is far more effective than re-reading theory.

These direct-knowledge questions test whether you remember key definitions and properties. Each is worth 1 mark but often appears in multiple-choice or fill-in-the-blank format. **Question 1:** If △ABC ~ △DEF and AB = 4 cm, DE = 6 cm, find the ratio of their areas. *Answer:* Area(△ABC)/Area(△DEF) = (AB/DE)² = (4/6)² = 16/36 = 4/9 *Key point:* The ratio of areas of similar triangles equals the square of the ratio of corresponding sides. **Question 2:** In △PQR, if a line parallel to side QR intersects PQ at X and PR at Y, what can you say about the triangles PXY and PQR? *Answer:* △PXY ~ △PQR (by AA similarity: ∠P is common, ∠PXY = ∠PQR as corresponding angles with XY ∥ QR). *Key point:* This is the Basic Proportionality Theorem (BPT) in action—always yields similar triangles. **Question 3:** Two triangles have a similarity ratio of 2:5. If the area of the smaller triangle is 32 cm², what is the area of the larger triangle? *Answer:* If linear ratio is 2:5, area ratio is 4:25. Let area of larger = A. Then 32/A = 4/25, so A = 32 × 25/4 = 200 cm². *Key point:* Always square the linear ratio to get the area ratio. **Question 4:** In △ABC, D and E are points on AB and AC such that DE ∥ BC. If AD = 3 cm, DB = 2 cm, AE = 4.5 cm, find EC. *Answer:* By BPT (Basic Proportionality Theorem): AD/DB = AE/EC, so 3/2 = 4.5/EC, EC = (4.5 × 2)/3 = 3 cm. *Key point:* BPT allows you to find unknown lengths without calculating the full length of a side. **Question 5:** State the SSS (Side-Side-Side) similarity criterion for triangles. *Answer:* If the sides of one triangle are proportional to the sides of another triangle, then the two triangles are similar. That is, if AB/DE = BC/EF = CA/FD, then △ABC ~ △DEF. *Key point:* This criterion requires you to check all three sides; if even one ratio differs, the triangles are not similar by SSS.

3-mark questions demand a short proof or a multi-step calculation. They often combine one theorem with a numerical application. **Question 1:** In △ABC, D and E are points on sides AB and AC respectively such that DE ∥ BC. If AB = 12 cm, AD = 8 cm, and AE = 6 cm, find AC. *Solution:* Given: DE ∥ BC, so by the Basic Proportionality Theorem, AD/DB = AE/EC We have AD = 8, AB = 12, so DB = AB − AD = 12 − 8 = 4. Let EC = x, so AC = AE + EC = 6 + x. Substitute into the BPT formula: 8/4 = 6/x 2 = 6/x x = 3 Therefore, AC = 6 + 3 = 9 cm. *Key point:* Always identify the parallel line first; it guarantees similarity and lets you apply BPT directly. **Question 2:** Prove that if two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. *Solution:* Let △ABC ~ △DEF with similarity ratio k = AB/DE = BC/EF = CA/FD. Since the triangles are similar, corresponding angles are equal. In particular, ∠A = ∠D. Area of △ABC = (1/2) × AB × AC × sin(∠A) Area of △DEF = (1/2) × DE × DF × sin(∠D) Since ∠A = ∠D and AC/DF = k (from similarity), Area(△ABC)/Area(△DEF) = (AB × AC)/(DE × DF) = (AB/DE) × (AC/DF) = k × k = k² Hence, the ratio of areas is the square of the similarity ratio. *Key point:* This proof uses the sine formula for area; examiners often accept the result directly if you state it clearly. **Question 3:** In △ABC, AB = 6 cm, BC = 8 cm, CA = 10 cm. In △PQR, PQ = 9 cm, QR = 12 cm, RP = 15 cm. Show that △ABC ~ △PQR and find the ratio of their areas. *Solution:* Check the ratios of corresponding sides: AB/PQ = 6/9 = 2/3 BC/QR = 8/12 = 2/3 CA/RP = 10/15 = 2/3 Since all three ratios are equal, by the SSS similarity criterion, △ABC ~ △PQR. The ratio of their areas is (2/3)² = 4/9. If area of △ABC = A and area of △PQR = A', then A/A' = 4/9. *Key point:* Always write the similarity statement in the correct order (matching vertices) to avoid errors. **Question 4:** In △ABC, ∠A = ∠D, AB/DE = AC/DF = 3/2. Prove that △ABC ~ △DEF. *Solution:* Given: ∠A = ∠D and AB/DE = AC/DF = 3/2. By the SAS (Side-Angle-Side) similarity criterion: if two sides of one triangle are proportional to two sides of another triangle, and the included angles are equal, then the triangles are similar. Here, AB and AC are proportional to DE and DF respectively (both in ratio 3/2), and the included angles ∠A and ∠D are equal. Therefore, △ABC ~ △DEF. *Key point:* SAS similarity requires the angle to be between the two proportional sides; if the angle is not included, the criterion does not apply. **Question 5:** The areas of two similar triangles are 36 cm² and 64 cm². If a side of the smaller triangle is 5 cm, find the corresponding side in the larger triangle. *Solution:* Let the similarity ratio be k. Then the ratio of areas is k². Area ratio = 36/64 = 9/16 So k² = 9/16, which gives k = 3/4. The corresponding side in the larger triangle is 5 ÷ k = 5 ÷ (3/4) = 5 × (4/3) = 20/3 ≈ 6.67 cm. Alternatively: if the side in the smaller triangle is 5, the corresponding side in the larger is 5/k = 5/(3/4) = 20/3 cm. *Key point:* When going from smaller to larger, divide the given side by k (the linear ratio of smaller to larger).

5-mark questions are multi-part problems combining proof, similarity, and calculation. They test deeper understanding and careful logical reasoning. **Question 1:** In △ABC, D is a point on AB such that AD/DB = 3/2, and E is a point on AC such that AE/EC = 3/2. Prove that DE ∥ BC and find DE in terms of BC. *Solution:* Given: AD/DB = 3/2 and AE/EC = 3/2. **Part 1: Prove DE ∥ BC** Since AD/DB = AE/EC, by the converse of the Basic Proportionality Theorem, DE ∥ BC. **Part 2: Find DE in terms of BC** Since DE ∥ BC, we have △ADE ~ △ABC (by AA similarity: ∠A is common, ∠ADE = ∠ABC as corresponding angles). From the similarity, AD/AB = DE/BC. Now, AD/DB = 3/2 means AD/(AD + DB) = 3/(3 + 2) = 3/5. So AD/AB = 3/5. Therefore, DE/BC = 3/5, which gives DE = (3/5)BC. **Key points:** The converse of BPT is crucial for proving parallelism; always use it when equal ratios are given. The similarity then gives the ratio of corresponding sides. --- **Question 2:** Triangles ABC and PQR are such that ∠B = ∠Q, AB/PQ = BC/QR. Prove that △ABC ~ △PQR. If AB = 6 cm, BC = 8 cm, PQ = 9 cm, find the area of △PQR given that the area of △ABC is 24 cm². *Solution:* **Part 1: Prove similarity** Given: ∠B = ∠Q and AB/PQ = BC/QR. Rearranging the side ratios: AB/PQ = BC/QR. Let this common ratio be k. Then AB = k·PQ and BC = k·QR. By the SAS similarity criterion (Side-Angle-Side): if two sides of one triangle are proportional to two sides of another, and the included angles are equal, the triangles are similar. Here, sides AB and BC (including angle B) are proportional to sides PQ and QR (including angle Q), with ∠B = ∠Q. Therefore, △ABC ~ △PQR. **Part 2: Find the area of △PQR** From AB/PQ = BC/QR: 6/9 = 8/QR QR = (8 × 9)/6 = 12 cm. The similarity ratio is k = AB/PQ = 6/9 = 2/3. The ratio of areas is k² = (2/3)² = 4/9. Area(△ABC)/Area(△PQR) = 4/9 24/Area(△PQR) = 4/9 Area(△PQR) = 24 × 9/4 = 54 cm². **Key points:** Always identify which sides are included in the given angle; only then can you apply SAS similarity. The area ratio follows directly from the similarity ratio squared. --- **Question 3:** In △ABC, D and E are the midpoints of sides AB and AC respectively. Prove that △ADE ~ △ABC and find the ratio of their perimeters and areas. If the perimeter of △ABC is 36 cm, find the perimeter of △ADE. *Solution:* **Part 1: Prove similarity** Given: D is the midpoint of AB, E is the midpoint of AC. So AD = DB = AB/2 and AE = EC = AC/2. Consider triangles ADE and ABC: - ∠A is common. - AD/AB = (AB/2)/AB = 1/2 - AE/AC = (AC/2)/AC = 1/2 By the SAS similarity criterion, △ADE ~ △ABC. **Part 2: Ratio of perimeters** When two triangles are similar with ratio k, their perimeters are also in ratio k. Here, k = AD/AB = 1/2. So Perimeter(△ADE)/Perimeter(△ABC) = 1/2. **Part 3: Ratio of areas** The ratio of areas is k² = (1/2)² = 1/4. **Part 4: Perimeter of △ADE** Given Perimeter(△ABC) = 36 cm. Perimeter(△ADE) = (1/2) × 36 = 18 cm. **Verification:** By the midpoint theorem, DE ∥ BC and DE = BC/2. So the third side of △ADE is also half the corresponding side of △ABC, confirming the 1:2 ratio for all sides and the perimeter. **Key points:** The midpoint theorem is a special case of BPT and always yields a similarity ratio of 1:2. Remember that the perimeter ratio equals the linear ratio, but the area ratio equals the square of the linear ratio.

The CBSE has gradually shifted the focus of Triangles questions in recent papers. Understanding these trends will help you prepare smartly: **Shift 1: More Applied & Real-World Contexts** Older papers (2015–2018) mostly asked pure geometric proofs. Recent papers (2022–2025) include indirect applications—e.g., 'A ladder leans against a wall, and a triangular support structure is placed. Use similar triangles to find the height.' Expect the 2026–27 papers to feature word problems blended with similarity and BPT. **Shift 2: Stronger Emphasis on Proof Verification** Instead of asking 'Prove △ABC ~ △DEF,' newer questions ask 'Show whether △ABC ~ △DEF given these measurements' or 'Which similarity criterion applies here and why?' This tests deeper conceptual understanding, not just memorized proofs. **Shift 3: Coordinate Geometry Integration** With the addition of coordinate geometry in Class 9 (Chapter 3), expect questions like 'Find the similarity ratio of two triangles given their vertices as coordinates.' This requires you to calculate side lengths using the distance formula, then apply similarity criteria. **Shift 4: Multi-Step Area Problems** Questions no longer stop at 'find the area ratio.' New papers ask 'If the area ratio is 4:9 and the perimeter of the larger triangle is P, find the perimeter of the smaller triangle' or involve finding actual areas using Heron's formula combined with similarity. **Shift 5: Less Emphasis on Bare Definitions** The number of 1-mark definition questions has dropped from ~3 per paper to ~1. Instead, examiners embed criterion names within multi-step problems, assuming you recognize them in context. **Strategy for 2026–27:** 1. Revise all four similarity criteria (AA, SSS, SAS, and the RHS criterion for right triangles) deeply, not just their statements. 2. Practice questions that mix coordinate geometry with similarity. 3. Work through word problems that require translating a real situation into a triangle diagram. 4. Solve area problems that combine Heron's formula (if given) with similarity ratios. 5. Practice writing concise, logical proofs without skipping steps—examiners mark harshly for unjustified leaps.

When you face a Triangles question in the board exam, follow this sequence to maximize marks: **Step 1: Identify the Type (15 seconds)** Read the question and determine whether it's asking for: - A proof of similarity or a parallel line - A calculation (side, area, or perimeter) - A statement or definition **Step 2: Spot Parallel Lines or Given Ratios (20 seconds)** If the problem mentions a line parallel to one side, immediately think BPT or converse BPT. If given equal ratios, the converse of BPT proves parallelism. This is the fastest path in most 3-mark questions. **Step 3: Choose the Right Criterion (20 seconds)** For similarity proofs: - If two angles are given or can be deduced (e.g., common angle, corresponding angles with parallel lines) → Use AA. - If all three sides (or their ratios) are given → Use SSS. - If two sides and the included angle are given (or can be deduced) → Use SAS. - For right triangles with hypotenuse and one side given → Use RHS. **Step 4: Write the Similarity Statement Clearly** Don't just say 'the triangles are similar.' Write: '△ABC ~ △DEF' (in correct vertex order). Examiners award marks for correctly identifying corresponding parts. **Step 5: For Area Questions, Always Square the Ratio** This is where many students slip up. If the linear ratio is 2:3, the area ratio is 4:9 (not 2:3 or 6:9). Write the formula once to be safe: Area(△₁)/Area(△₂) = [side₁/side₂]². **Step 6: Check Reasonableness** Before submitting: - Do the angles add up to 180° in any angle you've used? - Is the similarity ratio between the given measurements (e.g., if a side doubles, the similarity ratio should be > 1)? - Did you square the ratio when computing areas? **Timing Guide for a 5-Mark Question:** - Reading and identifying type: 1 min - Planning proof/approach: 1 min - Writing proof or calculation: 2 min - Reviewing and writing the final answer: 1 min Total: 5 minutes. This leaves buffer time for harder questions. **Common Pitfalls to Avoid:** 1. Forgetting the converse of BPT (if AD/DB = AE/EC, then DE ∥ BC). 2. Writing '△ABC ~ △PQR' when the actual similarity is '△ABC ~ △RPQ' (vertex order matters for identifying corresponding sides). 3. Confusing the similarity ratio with the area ratio; always square the linear ratio. 4. Stating a criterion without verifying all its conditions (e.g., claiming AA similarity when you've only shown one angle is equal). 5. Mixing up 'corresponding sides' in a similarity statement; always refer to your similarity notation. Start a 3-day free trial at cbsetutor.ai to get AI-guided solutions for every question in this chapter, with instant feedback on your proof-writing style and calculation accuracy.

Chapter 6: Triangles is central to Class 9 geometry and forms the foundation for higher mathematics. By working through the 13 previous year questions in this guide—5 one-mark, 5 three-mark, and 3 five-mark—you'll: 1. Internalize the four similarity criteria and recognize when to apply each. 2. Master the Basic Proportionality Theorem and its converse, the two most-repeated concepts. 3. Never confuse linear ratios with area ratios again. 4. Learn to write proofs in the exact style examiners expect. 5. Build confidence for unseen questions in your board exam. The key insight: similar triangles are not just a topic; they're a tool used across geometry, trigonometry, and even physics. Mastering them now ensures you understand related concepts in Classes 10, 11, and 12. **Next Steps:** - Solve each question independently without looking at the solution. - Mark your work; identify which criterion or theorem you missed if you went wrong. - Re-solve the question the next day from scratch. - Once you score 90%+ on all 13 questions, move to your textbook's exercise problems and other competitive exam materials. Do not rush. Consistency and understanding trump speed. Good luck with your practice!