Class 7 Maths Chapter 7 NCERT Solutions: A Tale of Three Intersecting Lines

NCERT Solutions for Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines (2025-26)

In-Text Questions and Activities

1. Construct a triangle in which all the sides are of length 4 cm.

Answer:

• Draw a line segment AB = 4 cm.
• With A as center and radius 4 cm, draw an arc.
• With B as center and radius 4 cm, draw another arc to intersect the previous arc at point C.
• Join AC and BC. ΔABC is the required equilateral triangle.

2. Construct a triangle of side lengths 4 cm, 5 cm, and 6 cm.

Answer:

• Draw a base AB = 4 cm.
• From A, draw an arc of radius 5 cm.
• From B, draw an arc of radius 6 cm to intersect the previous arc at C.
• Join AC and BC. The triangle ABC is the required triangle.

3. Construct triangles having the following side lengths (all the units are in cm):
(a) 4, 4, 6   (b) 3, 4, 5   (c) 1, 5, 5   (d) 4, 6, 8   (e) 3.5, 3.5, 3.5

Answer:

• For each set, follow this method: Draw the base equal to one side, construct arcs from each end with the other two side lengths as radii, mark intersection, and complete the triangle.
• For set (e), all sides are equal (3.5 cm), so the triangle formed will be equilateral.

4. Construct a triangle with side lengths 3 cm, 4 cm, and 8 cm. Is it possible?

Answer: No, it is not possible to construct a triangle with sides 3 cm, 4 cm, and 8 cm. The sum of the two smaller sides (3 + 4 = 7 cm) is less than the third side (8 cm), so it fails the triangle inequality.

5. Consider other lengths: 2 cm, 3 cm, and 6 cm. Can a triangle be constructed?

Answer: No, a triangle cannot be constructed. Here also, the sum of 2 cm and 3 cm is 5 cm, which is less than the third side 6 cm.

6. What is the triangle inequality?

Answer: Triangle inequality states that for three side lengths to form a triangle, the length of each side must be less than the sum of the other two sides. That is, for sides a, b, c: a < b + c, b < a + c, c < a + b.

7. Check if a triangle exists for each of the following sets of lengths:
(a) 1, 100, 100   (b) 3, 6, 9   (c) 1, 1, 5   (d) 5, 10, 12

Answer:

• (a) 1, 100, 100: 1 + 100 = 101 > 100, 100 + 1 = 101 > 100, 100 + 100 = 200 > 1. Triangle exists.
• (b) 3, 6, 9: 3 + 6 = 9 (not greater). Triangle cannot exist.
• (c) 1, 1, 5: 1 + 1 = 2 < 5. Triangle cannot exist.
• (d) 5, 10, 12: 5 + 10 = 15 > 12, 10 + 12 = 22 > 5, 5 + 12 = 17 > 10. Triangle exists.

8. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any side length? Justify your answer.

Answer: Yes, there exists an equilateral triangle with side lengths of 50 units (of any unit), or indeed any positive length. Since all sides are equal, the triangle inequality is always satisfied: 50 < 50 + 50 → 50 < 100 is true. Thus, an equilateral triangle can be constructed for any length > 0.

9. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as side lengths:

• (a) 1, 100
• (b) 5, 5
• (c) 3, 7

Answer:

• (a) Possible values: values greater than 99 and less than 101 (i.e., 99.1, 99.5, 99.9, 100, 100.5). Any length between 99 and 101 (excluding equality) will work.
• (b) Third side should be less than 10 and greater than 0 (but more than 0). Possible values: 1, 3, 5, 7, 9.
• (c) Third side must be greater than 4 and less than 10 (3 + 7 = 10, 7 – 3 = 4). Possible values: 4.1, 5, 6, 7, 8, 9.9.

10. Construct a triangle ABC with AB = 5 cm, AC = 4 cm and ∠A = 45°.

Answer:

• Draw AB = 5 cm.
• At A, construct ∠A = 45° using a protractor.
• From A on this ray, mark AC = 4 cm. Label C.
• Join BC. Triangle ABC is formed.

11. Construct a triangle ABC where AB = 5 cm, ∠A = 45°, and ∠B = 80°.

Answer:

• Draw AB = 5 cm.
• At A, draw ∠A = 45°.
• At B, draw ∠B = 80°, so that the arms of 45° at A and 80° at B meet at C.
• Join points to complete ΔABC.

12. Find the third angle of a triangle when two of the angles are 36°, 72°.

Answer: Sum of angles in a triangle = 180°. Third angle = 180° – (36° + 72°) = 180° – 108° = 72°.

13. Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70°, what would the third angle be?

Answer: Sum of three angles = 70° + 70° + 70° = 210°. But the sum must be 180°, so it is not possible. If two are 70°, the third is 180° – 140° = 40°.

14. What is the sum of the angles of any triangle?

Answer: The sum of the three angles in any triangle is always 180°.

15. What is an altitude of a triangle?

Answer: An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).

16. Classify the following triangles based on sides:
(a) All sides equal (b) Two sides equal (c) All sides unequal

Answer:

• (a) Equilateral triangle
• (b) Isosceles triangle
• (c) Scalene triangle

Key Concepts of Triangles and Their Construction

In NCERT Solutions for Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines, students learn how every triangle is formed by three line segments. Understanding triangle construction rules builds a strong foundation for tackling complex geometry questions.

This chapter emphasizes the importance of the triangle inequality property. Knowing when a triangle is possible for given side lengths can help you solve school tests and competitive exams with confidence.

Regular practice of exercise-based NCERT solutions will reinforce key geometry concepts. Always check if the sum of any two sides is greater than the third to ensure the triangle’s existence in every problem.