Class 9 Mathematics Chapter 7: A Tale of Three Intersecting Lines — Important Questions & Answers

Triangle geometry is no longer peripheral—it forms the backbone of CBSE Class 9 assessment. The 2024-25 rationalized curriculum emphasizes conceptual clarity over rote learning. Questions on Chapter 7 test three core competencies: (1) classification of triangles by sides (equilateral, isosceles, scalene) and angles (acute, right, obtuse), (2) application of the angle sum property (∠A + ∠B + ∠C = 180°) to solve for unknown angles, and (3) understanding the exterior angle property (an exterior angle equals the sum of non-adjacent interior angles). Additionally, the triangle inequality theorem—stating that the sum of any two sides must exceed the third—appears in 'can we form a triangle?' problems. Board papers increasingly favour multi-step reasoning questions where students must combine two or three properties. For example, a 5-mark question might ask: given two angles of a triangle and a condition about side lengths, determine the triangle type and find the third angle. Practising these 18 questions systematically prepares you to handle such composite problems confidently.

**Q1: If a triangle has sides 3 cm, 4 cm, and 5 cm, it is classified as:** (A) Equilateral (B) Isosceles (C) Scalene (D) Right-angled **Answer: (C) Scalene** Explanation: A scalene triangle has all three sides of different lengths. Sides 3, 4, 5 are all unequal, so it's scalene. (Note: 3² + 4² = 9 + 16 = 25 = 5², so it's also a right-angled triangle, but the primary classification here is scalene.) **Q2: The angle sum of any triangle is:** (A) 90° (B) 180° (C) 270° (D) 360° **Answer: (B) 180°** Explanation: The angle sum property states that the sum of all interior angles in a triangle is always 180°, regardless of the triangle's type or size. **Q3: An exterior angle of a triangle is 120°. The sum of the two non-adjacent interior angles is:** (A) 60° (B) 120° (C) 240° (D) 180° **Answer: (B) 120°** Explanation: By the exterior angle property, an exterior angle equals the sum of the two opposite (non-adjacent) interior angles. So the answer is 120°. **Q4: Which set of lengths can form a triangle?** (A) 2 cm, 3 cm, 6 cm (B) 5 cm, 5 cm, 10 cm (C) 4 cm, 5 cm, 6 cm (D) 1 cm, 2 cm, 5 cm **Answer: (C) 4 cm, 5 cm, 6 cm** Explanation: Apply triangle inequality. For (C): 4+5=9>6 ✓, 5+6=11>4 ✓, 4+6=10>5 ✓. For (A): 2+3=5<6 ✗. For (B): 5+5=10≮10 ✗. For (D): 1+2=3<5 ✗. **Q5: A triangle with angles 60°, 60°, and 60° is:** (A) Isosceles acute-angled (B) Equilateral acute-angled (C) Scalene obtuse-angled (D) Right-angled **Answer: (B) Equilateral acute-angled** Explanation: Equal angles (60° each) mean equal opposite sides, making it equilateral. All angles < 90°, so it's acute-angled.

**Q1: Two angles of a triangle are 65° and 55°. Find the third angle and classify the triangle by angles.** **Solution:** Using angle sum property: ∠1 + ∠2 + ∠3 = 180° 65° + 55° + ∠3 = 180° ∠3 = 180° − 120° = 60° Since all angles are less than 90°, the triangle is **acute-angled**. **Q2: The sides of a triangle are in the ratio 3:4:5. If the perimeter is 36 cm, find the length of each side.** **Solution:** Let sides be 3x, 4x, 5x. Perimeter = 3x + 4x + 5x = 12x = 36 x = 3 Sides are 3(3)=9 cm, 4(3)=12 cm, 5(3)=15 cm. Verify: 9+12=21>15 ✓, 12+15=27>9 ✓, 9+15=24>12 ✓ **Q3: An exterior angle of a triangle is 110°. One of the non-adjacent interior angles is 50°. Find the other non-adjacent interior angle.** **Solution:** Let the two non-adjacent interior angles be 50° and x°. By exterior angle property: 50° + x° = 110° x = 60° The other non-adjacent interior angle is **60°**. **Q4: A triangle has sides 6 cm, 8 cm, and 10 cm. Is it a right-angled triangle? Justify.** **Solution:** Check if sides satisfy the Pythagorean theorem: a² + b² = c² 6² + 8² = 36 + 64 = 100 10² = 100 Since 6² + 8² = 10², **yes, it is a right-angled triangle** with the right angle opposite the 10 cm side. **Q5: In a triangle, one angle is 90° and the other two angles are equal. Find these two angles and classify the triangle by sides.** **Solution:** Let the two equal angles be x° each. x + x + 90° = 180° 2x = 90° x = 45° The two equal angles are **45° each**. Since two angles are equal, the opposite sides are equal, making it **isosceles right-angled triangle**.

**Q1: The exterior angles of a triangle are (2x+10)°, (3x+20)°, and (4x+30)°. Find the value of x and each exterior angle.** **Solution:** Sum of exterior angles of any polygon is 360°. (2x+10)° + (3x+20)° + (4x+30)° = 360° 9x + 60 = 360 9x = 300 x = 300/9 = 100/3 ≈ 33.33° Exterior angles: 2(100/3)+10 = 200/3 + 10 = 230/3 ≈ 76.67° 3(100/3)+20 = 100 + 20 = 120° 4(100/3)+30 = 400/3 + 30 = 490/3 ≈ 163.33° Verify: 230/3 + 120 + 490/3 = (230+490)/3 + 120 = 720/3 + 120 = 240 + 120 = 360° ✓ **Q2: In triangle ABC, ∠B = 90°, ∠A = 35°. An exterior angle at C is drawn. Find (i) ∠C, (ii) the exterior angle at C, and (iii) verify using the exterior angle property.** **Solution:** (i) Using angle sum: ∠A + ∠B + ∠C = 180° 35° + 90° + ∠C = 180° ∠C = 55° (ii) Interior angle C and its exterior angle are supplementary: Exterior angle at C = 180° − 55° = 125° (iii) Verify: Exterior angle at C = ∠A + ∠B 125° = 35° + 90° = 125° ✓ **Q3: Three sides of a triangle are (a+1) cm, (a+2) cm, and (a+3) cm, where a > 0. For what values of a can these form a valid triangle?** **Solution:** Apply triangle inequality for all three conditions: (1) (a+1) + (a+2) > (a+3) 2a + 3 > a + 3 a > 0 ... (given) (2) (a+2) + (a+3) > (a+1) 2a + 5 > a + 1 a > −4 (always true for a > 0) (3) (a+1) + (a+3) > (a+2) 2a + 4 > a + 2 a > −2 (always true for a > 0) Combining all: **a > 0** (the given constraint is the binding one). So for any a > 0, these sides form a valid triangle. **Q4: In triangle PQR, ∠P = ∠Q = 75°. Classify the triangle by sides and angles. Also, find ∠R and an exterior angle at R.** **Solution:** Since ∠P = ∠Q = 75°, two angles are equal. Using angle sum: ∠P + ∠Q + ∠R = 180° 75° + 75° + ∠R = 180° ∠R = 30° Classification by angles: One angle (30°) is acute, two angles (75° each) are acute. **All angles < 90°, so it's acute-angled**. Classification by sides: Since ∠P = ∠Q, the opposite sides QR = PR. **It's isosceles**. Exterior angle at R = 180° − 30° = **150°** Verify: Exterior at R = ∠P + ∠Q = 75° + 75° = 150° ✓

**Q1: In triangle ABC, the exterior angle at vertex C is 130°. The interior angles ∠A and ∠B are in the ratio 3:2. Find all three interior angles and classify the triangle by angles.** **Solution:** Let ∠A = 3k and ∠B = 2k, where k is a positive constant. By the exterior angle property: Exterior angle at C = ∠A + ∠B 130° = 3k + 2k 130° = 5k k = 26° Therefore: ∠A = 3(26°) = 78° ∠B = 2(26°) = 52° Using angle sum property: ∠A + ∠B + ∠C = 180° 78° + 52° + ∠C = 180° ∠C = 180° − 130° = 50° Verify exterior angle: ∠C (interior) = 180° − 130° = 50° ✓ Alternate check: 78° + 52° + 50° = 180° ✓ **Classification:** All three angles (78°, 52°, 50°) are less than 90°, so the triangle is **acute-angled**. All three sides are different (since all angles are different), so it's also **scalene**. **Answer: ∠A = 78°, ∠B = 52°, ∠C = 50°; Triangle is acute-angled scalene.** --- **Q2: The perimeter of a triangle is 50 cm. The sides are (x+5) cm, (x+10) cm, and (2x−5) cm. Find the length of each side, verify the triangle inequality, and classify by sides.** **Solution:** Perimeter = (x+5) + (x+10) + (2x−5) = 50 4x + 10 = 50 4x = 40 x = 10 Sides are: Side 1: 10 + 5 = 15 cm Side 2: 10 + 10 = 20 cm Side 3: 2(10) − 5 = 15 cm **Triangle Inequality Check:** (1) 15 + 20 = 35 > 15 ✓ (2) 20 + 15 = 35 > 15 ✓ (3) 15 + 15 = 30 > 20 ✓ All conditions satisfied; valid triangle. **Classification by Sides:** Since two sides are equal (15 cm each) and one is different (20 cm), the triangle is **isosceles**. **Additional Check (by angles):** Using the cosine rule to find the angle opposite the 20 cm side: c² = a² + b² − 2ab cos C 20² = 15² + 15² − 2(15)(15) cos C 400 = 225 + 225 − 450 cos C 400 = 450 − 450 cos C 450 cos C = 50 cos C = 50/450 = 1/9 ≈ 0.111 C ≈ 83.6° (acute) The other two equal angles: ∠A = ∠B = (180° − 83.6°)/2 = 48.2° (acute) All angles are acute, so it's **acute-angled isosceles triangle**. **Answer: Sides are 15 cm, 20 cm, 15 cm; Triangle is isosceles acute-angled.** --- **Q3: A triangle has angles in the ratio 2:3:4. Find each angle, determine the type of triangle by angles, and find the exterior angle at the largest interior angle.** **Solution:** Let the angles be 2m, 3m, and 4m. Using angle sum property: 2m + 3m + 4m = 180° 9m = 180° m = 20° Angles are: ∠A = 2(20°) = 40° ∠B = 3(20°) = 60° ∠C = 4(20°) = 80° **Verification:** 40° + 60° + 80° = 180° ✓ **Classification by Angles:** All angles (40°, 60°, 80°) are less than 90°, so the triangle is **acute-angled**. **Largest Interior Angle:** ∠C = 80° is the largest. **Exterior Angle at C:** Exterior angle = 180° − 80° = 100° Alternate verification (exterior angle property): Exterior angle at C = ∠A + ∠B = 40° + 60° = 100° ✓ **Answer: Angles are 40°, 60°, 80°; Triangle is acute-angled; Exterior angle at largest angle = 100°.**

**Case Study: Road Safety Triangle Markers** A road construction company places reflective triangular markers along a mountain highway. Each marker is a physical triangle with sides meeting at three vertices. Quality control requires: - All triangles must have a perimeter of 24 cm. - No two sides can be equal (scalene requirement for distinctive appearance). - Each angle must be different, and the largest angle cannot exceed 120° (for structural stability). The company manufactures a batch with sides (a), (a+2), and (a+4) cm, where a is a positive integer. **Sub-questions:** **(i) For what integer values of a do these sides form a valid triangle?** **Solution:** Perimeter: a + (a+2) + (a+4) = 24 3a + 6 = 24 3a = 18 a = 6 cm Sides are 6 cm, 8 cm, 10 cm. Triangle inequality check: (1) 6 + 8 = 14 > 10 ✓ (2) 8 + 10 = 18 > 6 ✓ (3) 6 + 10 = 16 > 8 ✓ **Answer: a = 6 cm (unique solution for perimeter 24 cm).** **(ii) Verify that this triangle is scalene.** **Solution:** Sides are 6 cm, 8 cm, 10 cm. All three are different (6 ≠ 8 ≠ 10). **Yes, the triangle is scalene.** ✓ **(iii) Check if this triangle meets the structural stability condition (largest angle ≤ 120°).** **Solution:** Note: 6² + 8² = 36 + 64 = 100 = 10² By the converse of Pythagorean theorem, this is a **right-angled triangle** with the right angle opposite the 10 cm side. So the largest angle is 90°. Since 90° < 120°, **the triangle meets the structural stability condition.** ✓ **(iv) Find all three angles and classify the triangle by angles.** **Solution:** The triangle is right-angled, so one angle = 90°. Let the other two angles be ∠A and ∠B. ∠A + ∠B = 180° − 90° = 90° Using sine rule or basic trigonometry: sin ∠A = opposite/hypotenuse = 6/10 = 0.6 ∠A = arcsin(0.6) ≈ 36.87° ≈ 37° sin ∠B = 8/10 = 0.8 ∠B = arcsin(0.8) ≈ 53.13° ≈ 53° Verification: 37° + 53° + 90° = 180° ✓ **Classification: Right-angled scalene triangle.** Angles (in order): 37°, 53°, 90°; all different. Sides: 6 cm, 8 cm, 10 cm; all different. **(v) The company wants to design a variant with sides in ratio 3:4:5. Will this variant always meet both the scalene and stability conditions regardless of size?** **Solution:** Any triangle with sides in ratio 3:4:5 is a right-angled triangle (since 3² + 4² = 5²). So the largest angle = 90° < 120° ✓ (meets stability). For the triangle to be scalene, the sides 3k, 4k, 5k must all be different. Since 3k ≠ 4k ≠ 5k for any k > 0, the triangle is always scalene. ✓ **Answer: Yes, any 3:4:5 variant will always meet both conditions, regardless of scale (as long as k > 0).** --- **Overall Case-Study Conclusion:** The markers with sides 6 cm, 8 cm, 10 cm are an excellent choice. They satisfy the perimeter requirement (24 cm), are scalene, have all different angles (37°, 53°, 90°), and meet the structural stability rule. This case-study integrates properties of triangles—classification, angle sum, Pythagorean theorem, and triangle inequality—into a real-world scenario.

At cbsetutor.ai, our AI tutor is engineered to mimic the classroom teaching style of experienced CBSE mathematics tutors, with a focus on deep comprehension and exam readiness. Here's exactly how we drill Chapter 7 concepts: **Adaptive Question Generation:** Our AI generates unlimited variations of every question type above. For instance, if you solve an angle sum problem with angles in ratio 2:3:5, the next drill generates a new ratio (3:4:5, 1:2:3, etc.) with different triangle constraints. This prevents memorization and forces genuine understanding. **Step-by-Step Walkthroughs:** Every solution is broken into digestible steps. When you solve incorrectly, the AI doesn't just show the answer—it identifies precisely where your logic broke down. For example: "You correctly applied angle sum, but didn't recognize this is an isosceles triangle because two angles are equal." **Concept Linking:** The tutor connects all four topics—triangle types, angle sum, exterior angle, and triangle inequality—into unified problem-solving. A single 5-mark question might require you to (1) check triangle inequality, (2) find the third angle, (3) identify if it's isosceles or scalene, and (4) compute an exterior angle. We guide you through all four steps without hand-holding. **Spaced Repetition on Weak Areas:** If you struggle with exterior angle property, the AI increases the frequency of those drills in your weekly schedule. Conversely, if you master angle classification, we reduce repetition and move to harder HOTS questions. **Board Exam Simulation:** Every Friday, our tutors set a 30-minute mock test with the exact paper structure of CBSE Class 9 board exams (1-mark, 2-mark, 3-mark, 5-mark sections). You solve under time pressure, and the AI provides performance analytics: which question types you're fast at, which require more thinking time, and where careless errors creep in. **Live Doubt Clearing:** Beyond AI drills, CBSETUTOR.ai connects you with expert human tutors (available 6 PM–9 PM IST daily) if an AI explanation doesn't click. This hybrid model ensures you never get stuck. **Glossary & Quick Recall Cards:** We maintain a flashcard deck of 20 key formulas and theorems from Chapter 7 (angle sum = 180°, exterior angle = sum of opposite interior angles, etc.). Daily 5-minute reviews keep these locked in memory. Start your free 3-day trial today and experience how AI-driven daily drilling transforms your Chapter 7 mastery into 90%+ board exam scores.