Class 7 : Maths – Lesson 7. A Tale of Three Intersecting Lines

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On December 11, 2025

EXPLANATION AND ANALYSIS

🔵 Introduction: When Lines Meet in Different Ways

🧠 In earlier lessons, we learned about parallel lines and intersecting lines. We saw how two lines can meet at a point and form angles. In real life and geometry, there are situations where three lines intersect with each other, creating interesting patterns and relationships between angles.

🌿 Examples from daily life
🔵 Road junctions with three roads
🟢 The letter “Y” shape
🟡 Triangular supports in bridges
🔴 Arrow signs and direction boards

This chapter explores how three intersecting lines behave and what special angle relationships are formed when they meet.

🟢 Understanding Intersecting Lines Again

🧠 Intersecting lines are lines that meet or cross at a point.

🔹 Two intersecting lines meet at one point
🔹 They form angles at the point of intersection

📌 Example
Two roads crossing at a junction are intersecting lines.

💡 Concept:
Angles are formed only when lines intersect.

🔵 Three Intersecting Lines

🧠 When three lines intersect at a single point, they divide the space around the point into several angles.

🔹 All three lines pass through the same point
🔹 More angles are formed compared to two intersecting lines

📌 Example
Three sticks tied together at one end form three intersecting lines.

✏️ Note:
All three lines must meet at the same point to be called three intersecting lines.

🟢 Angles Formed by Three Intersecting Lines

🧠 Three intersecting lines form six angles at the point of intersection.

🔹 These angles are arranged around the point
🔹 Some angles may be equal
🔹 All angles together make a full turn

📌 Observation
The sum of all angles around a point is 360 degrees.

💡 Concept:
This rule helps us find unknown angles easily.

🔵 Adjacent Angles

🧠 Adjacent angles are angles that lie next to each other.

🔹 They share a common arm
🔹 They share a common vertex
🔹 They do not overlap

📌 Example
Two angles formed side by side by intersecting lines are adjacent angles.

✏️ Note:
Adjacent angles together form a larger angle.

🟢 Vertically Opposite Angles

🧠 Vertically opposite angles are angles formed when two lines intersect.

🔹 They lie opposite to each other
🔹 They are always equal in measure

📌 Example
If one angle is 40°, the vertically opposite angle is also 40°.

💡 Concept:
This property holds true even when more lines pass through the same point.

🔵 Angle Sum Around a Point

🧠 When several angles meet at a point, their total sum is always 360 degrees.

🔹 This is called a complete angle
🔹 It represents a full rotation

📌 Example
If five angles around a point are known, the sixth can be found by subtracting their sum from 360°.

✏️ Note:
This rule is very important for solving angle problems.

🟡 Special Angle Relationships with Three Lines

🧠 When three lines intersect, different types of angle relationships appear.

🔹 Adjacent angles
🔹 Vertically opposite angles
🔹 Angles around a point

📌 Example
In a three-line intersection, opposite angles across the same two lines remain equal.

💡 Concept:
Understanding these relationships helps in solving geometry problems.

🔴 Common Mistakes to Avoid

🔴 Forgetting that all angles around a point add up to 360°
🔴 Confusing adjacent angles with vertically opposite angles
🔴 Assuming all angles formed are equal
🔴 Ignoring the position of lines

✏️ Note:
Always draw a neat figure and label angles clearly.

🟢 Importance of Studying Three Intersecting Lines

🧠 This chapter helps students to:

🔹 Understand angle relationships clearly
🔹 Solve geometry questions confidently
🔹 Build a strong base for transversals and polygons
🔹 Apply geometry in real-life designs

These ideas are essential for advanced geometry in higher classes.

📘 Summary

🔵 Three intersecting lines meet at a single point
🟢 They form six angles
🟡 Vertically opposite angles are equal
🔴 Adjacent angles lie next to each other
🔵 Sum of angles around a point is 360°
🟢 Angle relationships help find unknown angles

📝 Quick Recap

📝 Quick Recap
🔵 Intersecting lines meet at a point
🟢 Three intersecting lines form six angles
🟡 Vertically opposite angles are equal
🔴 Adjacent angles share a common side
🔵 Angles around a point add up to 360°

——————————————————————————————————————————————————————————————————————————–

TEXTBOOK QUESTIONS

🔵 1. FIGURE IT OUT ?

🔒 ❓ 1. Use the points on the circle and/or the centre to form isosceles triangles.

📌 ✅ Answer:
🔹 Let the centre be O and take any two distinct points on the circle, say P and Q.
🔹 Join OP, OQ, and PQ to form triangle OPQ.
🔹 OP = OQ because both are radii of the same circle.
🔹 Therefore, triangle OPQ is isosceles.

🔹 More isosceles triangles (same idea):
🔸 Choose any other pair of points R and S on the circle.
🔸 Join OR, OS, and RS.
🔸 OR = OS (radii), so triangle ORS is isosceles.

🔒 ❓ 2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.

📌 ✅ Answer:

🔹 First diagram (two equal circles with centres A and B):
🔸 Let the circles intersect at two points; call them P and Q.
🔸 Join A to P and A to Q, and join P to Q.
🔸 AP = AQ (radii of circle with centre A), so triangle APQ is isosceles.
🔸 Similarly, join B to P and B to Q, and join P to Q.
🔸 BP = BQ (radii of circle with centre B), so triangle BPQ is isosceles.

🔹 Second diagram (three equal circles with centres A, B, and C):
🔸 A, B, and C are centres of equal circles.
🔸 From the diagram, each centre lies on the other two circles.
🔸 So AB is a radius (A lies on B’s circle and B is its centre), hence AB = r.
🔸 Similarly, BC = r and CA = r.
🔸 Therefore AB = BC = CA.
🔹 So triangle ABC is equilateral.

🔹 Also, isosceles triangles can be formed using intersection points:
🔸 Take any two intersection points on the circle with centre A, say P and Q.
🔸 Then AP = AQ (radii), so triangle APQ is isosceles.
🔸 Same works with centre B or centre C.

🔵 2. FIGURE IT OUT ?

🔒 ❓ 1. We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

📌 ✅ Answer:
🔹 A triangle exists only if the sum of any two sides is greater than the third side.

🔹 For 3 cm, 4 cm, 8 cm:
🔸 3 + 4 = 7
🔸 7 < 8
🔹 Since the sum of two sides is less than the third side, triangle cannot be formed.

🔹 For 2 cm, 3 cm, 6 cm:
🔸 2 + 3 = 5
🔸 5 < 6
🔹 Again, the sum of two sides is less than the third side.
🔹 So no triangle is possible.

🔹 Therefore, without construction, using the triangle inequality rule, we can conclude that such triangles do not exist.

🔒 ❓ 2. Can we say anything about the existence of a triangle for each of the following sets of lengths?

🔒 ❓ (a) 10 km, 10 km and 25 km

📌 ✅ Answer:
🔹 10 + 10 = 20
🔹 20 < 25
🔹 The sum of two sides is less than the third side.
🔹 Therefore, no triangle can be formed.

🔒 ❓ (b) 5 mm, 10 mm and 20 mm

📌 ✅ Answer:
🔹 5 + 10 = 15
🔹 15 < 20
🔹 The triangle inequality condition is not satisfied.
🔹 So, no triangle is possible.

🔒 ❓ (c) 12 cm, 20 cm and 40 cm

📌 ✅ Answer:
🔹 12 + 20 = 32
🔹 32 < 40
🔹 The sum of two sides is less than the third side.
🔹 Therefore, triangle cannot be formed.

🔒 ❓ 3. For the set of lengths 10 cm, 15 cm and 30 cm, check the comparisons.

📌 ✅ Answer:
🔹 10 < 15 + 30
🔸 10 < 45 (true)

🔹 15 < 10 + 30
🔸 15 < 40 (true)

🔹 30 < 10 + 15
🔸 30 < 25 (false)

🔹 Since one comparison fails (30 is greater than the sum of the other two), triangle cannot be formed.

🔹 Conclusion: For a triangle to exist, the sum of any two sides must be strictly greater than the third side in all three comparisons.

🔵 3. FIGURE IT OUT ?

🔒 ❓ 1. Which of the following lengths can be the sidelengths of a triangle? Explain your answers.

📌 ✅ Answer:
🔹 A triangle is possible only if the sum of any two sides is greater than the third side.
🔹 We check each set using the triangle inequality.

🔒 ❓ (a) 2, 2, 5

📌 ✅ Answer:
🔹 2 + 2 = 4
🔹 4 < 5
🔹 Condition not satisfied.
🔹 Triangle not possible.

🔒 ❓ (b) 3, 4, 6

📌 ✅ Answer:
🔹 3 + 4 = 7 > 6
🔹 3 + 6 = 9 > 4
🔹 4 + 6 = 10 > 3
🔹 All three conditions satisfied.
🔹 Triangle is possible.

🔒 ❓ (c) 2, 4, 8

📌 ✅ Answer:
🔹 2 + 4 = 6
🔹 6 < 8
🔹 Condition fails.
🔹 Triangle not possible.

🔒 ❓ (d) 5, 5, 8

📌 ✅ Answer:
🔹 5 + 5 = 10 > 8
🔹 5 + 8 = 13 > 5
🔹 5 + 8 = 13 > 5
🔹 All conditions satisfied.
🔹 Triangle is possible.

🔒 ❓ (e) 10, 20, 25

📌 ✅ Answer:
🔹 10 + 20 = 30 > 25
🔹 10 + 25 = 35 > 20
🔹 20 + 25 = 45 > 10
🔹 All conditions satisfied.
🔹 Triangle is possible.

🔒 ❓ (f) 10, 20, 35

📌 ✅ Answer:
🔹 10 + 20 = 30
🔹 30 < 35
🔹 Condition fails.
🔹 Triangle not possible.

🔒 ❓ (g) 24, 26, 28

📌 ✅ Answer:
🔹 24 + 26 = 50 > 28
🔹 24 + 28 = 52 > 26
🔹 26 + 28 = 54 > 24
🔹 All three conditions satisfied.
🔹 Triangle is possible.

📌 ✅ Final Conclusion:
🔹 A triangle can be formed for sets (b), (d), (e), and (g).
🔹 A triangle cannot be formed for sets (a), (c), and (f).

🔵 4. FIGURE IT OUT ?

🔒 ❓ 1. Check if a triangle exists for each of the following set of lengths:

🔒 ❓ (a) 1, 100, 100

📌 ✅ Answer:
🔹 1 + 100 = 101 > 100
🔹 100 + 100 = 200 > 1
🔹 1 + 100 = 101 > 100
🔹 All three conditions satisfied.
🔹 Triangle exists.

🔒 ❓ (b) 3, 6, 9

📌 ✅ Answer:
🔹 3 + 6 = 9
🔹 9 is not greater than 9.
🔹 Triangle inequality not satisfied.
🔹 Triangle does not exist.

🔒 ❓ (c) 1, 1, 5

📌 ✅ Answer:
🔹 1 + 1 = 2
🔹 2 < 5
🔹 Condition fails.
🔹 Triangle does not exist.

🔒 ❓ (d) 5, 10, 12

📌 ✅ Answer:
🔹 5 + 10 = 15 > 12
🔹 5 + 12 = 17 > 10
🔹 10 + 12 = 22 > 5
🔹 All conditions satisfied.
🔹 Triangle exists.

🔒 ❓ 2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.

📌 ✅ Answer:
🔹 For sides 50, 50, 50:
🔸 50 + 50 = 100 > 50
🔸 Condition satisfied.
🔹 Therefore, such an equilateral triangle exists.

🔹 In general:
🔸 For any positive number a,
🔸 a + a = 2a > a
🔸 Hence, triangle inequality is always satisfied.
🔹 So an equilateral triangle exists for any positive sidelength.

🔒 ❓ 3. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths:

🔒 ❓ (a) 1, 100

📌 ✅ Answer:
🔹 Third side x must satisfy:
🔸 100 − 1 < x < 100 + 1
🔸 99 < x < 101
🔹 Five possible values:
🔸 99.2
🔸 99.5
🔸 100
🔸 100.4
🔸 100.8

🔒 ❓ (b) 5, 5

📌 ✅ Answer:
🔹 Third side x must satisfy:
🔸 5 − 5 < x < 5 + 5
🔸 0 < x < 10
🔹 Five possible values:
🔸 2
🔸 4
🔸 6
🔸 8
🔸 9

🔒 ❓ (c) 3, 7

📌 ✅ Answer:
🔹 Third side x must satisfy:
🔸 7 − 3 < x < 7 + 3
🔸 4 < x < 10
🔹 Five possible values:
🔸 5
🔸 6
🔸 7
🔸 8
🔸 9

📌 ✅ Final Observation:
🔹 For two given sides a and b, the third side must satisfy
🔸 |a − b| < x < a + b

🔵 5. FIGURE IT OUT ?

🔒 ❓ 1. Construct triangles for the following measurements where the angle is included between the sides:

🔒 ❓ (a) 3 cm, 75°, 7 cm

📌 ✅ Answer:
🔹 Step 1: Draw a line segment AB = 7 cm.
🔹 Step 2: At point A, construct an angle of 75°.
🔹 Step 3: On the second arm of the angle, mark point C such that AC = 3 cm.
🔹 Step 4: Join BC.
🔹 Triangle ABC is the required triangle.

🔒 ❓ (b) 6 cm, 25°, 3 cm

📌 ✅ Answer:
🔹 Step 1: Draw a line segment PQ = 6 cm.
🔹 Step 2: At point P, construct an angle of 25°.
🔹 Step 3: On the second arm of the angle, mark point R such that PR = 3 cm.
🔹 Step 4: Join QR.
🔹 Triangle PQR is the required triangle.

🔒 ❓ (c) 3 cm, 120°, 8 cm

📌 ✅ Answer:
🔹 Step 1: Draw a line segment XY = 8 cm.
🔹 Step 2: At point X, construct an angle of 120°.
🔹 Step 3: On the second arm of the angle, mark point Z such that XZ = 3 cm.
🔹 Step 4: Join YZ.
🔹 Triangle XYZ is the required triangle.

📌 ✅ Observation:
🔹 In each case, two sides and the included angle are given.
🔹 A triangle can be uniquely constructed using the SAS (Side–Angle–Side) condition.

🔵 6. FIGURE IT OUT ?

🔒 ❓ 1. Construct triangles for the following measurements:

🔒 ❓ (a) 75°, 5 cm, 75°

📌 ✅ Answer:
🔹 Step 1: Draw a line segment AB = 5 cm.
🔹 Step 2: At point A, construct an angle of 75°.
🔹 Step 3: At point B, construct an angle of 75° on the same side of AB.
🔹 Step 4: Let the two arms meet at point C.
🔹 Triangle ABC is the required triangle.
🔹 The third angle will be 180° − (75° + 75°) = 30°.

🔒 ❓ (b) 25°, 3 cm, 60°

📌 ✅ Answer:
🔹 Step 1: Draw a line segment PQ = 3 cm.
🔹 Step 2: At point P, construct an angle of 25°.
🔹 Step 3: At point Q, construct an angle of 60° on the same side of PQ.
🔹 Step 4: Let the two arms meet at point R.
🔹 Triangle PQR is the required triangle.
🔹 The third angle will be 180° − (25° + 60°) = 95°.

🔒 ❓ (c) 120°, 6 cm, 30°

📌 ✅ Answer:
🔹 Step 1: Draw a line segment XY = 6 cm.
🔹 Step 2: At point X, construct an angle of 120°.
🔹 Step 3: At point Y, construct an angle of 30° on the same side of XY.
🔹 Step 4: Let the two arms meet at point Z.
🔹 Triangle XYZ is the required triangle.
🔹 The third angle will be 180° − (120° + 30°) = 30°.

📌 ✅ Observation:
🔹 In each case, two angles and the included side are given.
🔹 A triangle can be uniquely constructed using the ASA (Angle–Side–Angle) condition.

🔵 7. FIGURE IT OUT ?

🔒 ❓ 1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category:

🔒 ❓ (a) 30°

📌 ✅ Answer:
🔹 In a triangle, sum of angles = 180°.

🔹 (a) Triangle possible:
🔸 Choose 60° → 30° + 60° = 90° < 180° ✔
🔸 Choose 100° → 30° + 100° = 130° < 180° ✔

🔹 (b) Triangle not possible:
🔸 Choose 150° → 30° + 150° = 180° ✘
🔸 Choose 160° → 30° + 160° = 190° ✘

🔒 ❓ (b) 70°

📌 ✅ Answer:
🔹 (a) Triangle possible:
🔸 50° → 70° + 50° = 120° < 180° ✔
🔸 80° → 70° + 80° = 150° < 180° ✔

🔹 (b) Triangle not possible:
🔸 110° → 70° + 110° = 180° ✘
🔸 120° → 70° + 120° = 190° ✘

🔒 ❓ (c) 54°

📌 ✅ Answer:
🔹 (a) Triangle possible:
🔸 60° → 54° + 60° = 114° < 180° ✔
🔸 100° → 54° + 100° = 154° < 180° ✔

🔹 (b) Triangle not possible:
🔸 126° → 54° + 126° = 180° ✘
🔸 140° → 54° + 140° = 194° ✘

🔒 ❓ (d) 144°

📌 ✅ Answer:
🔹 (a) Triangle possible:
🔸 20° → 144° + 20° = 164° < 180° ✔
🔸 30° → 144° + 30° = 174° < 180° ✔

🔹 (b) Triangle not possible:
🔸 36° → 144° + 36° = 180° ✘
🔸 50° → 144° + 50° = 194° ✘

🔒 ❓ 2. Determine which of the following pairs can be the angles of a triangle and which cannot:

🔒 ❓ (a) 35°, 150°

📌 ✅ Answer:
🔹 35° + 150° = 185° > 180°
🔹 Not possible.

🔒 ❓ (b) 70°, 30°

📌 ✅ Answer:
🔹 70° + 30° = 100° < 180°
🔹 Possible.
🔹 Third angle = 180° − 100° = 80°.

🔒 ❓ (c) 90°, 85°

📌 ✅ Answer:
🔹 90° + 85° = 175° < 180°
🔹 Possible.
🔹 Third angle = 5°.

🔒 ❓ (d) 50°, 150°

📌 ✅ Answer:
🔹 50° + 150° = 200° > 180°
🔹 Not possible.

📌 ✅ Final Rule:
🔹 For two angles to form a triangle, their sum must be strictly less than 180°.

🔵 8. FIGURE IT OUT ?

🔒 ❓ 1. Find the third angle of a triangle (using a parallel line) when two of the angles are:

🔒 ❓ (a) 36°, 72°

📌 ✅ Answer:
🔹 In a triangle, sum of angles = 180°.
🔹 Third angle = 180° − (36° + 72°)
🔹 Third angle = 180° − 108°
🔹 Third angle = 72°

✔️ Final Answer: 72°

🔒 ❓ (b) 150°, 15°

📌 ✅ Answer:
🔹 Sum of angles in triangle = 180°.
🔹 Third angle = 180° − (150° + 15°)
🔹 Third angle = 180° − 165°
🔹 Third angle = 15°

✔️ Final Answer: 15°

🔒 ❓ (c) 90°, 30°

📌 ✅ Answer:
🔹 Sum of the two given angles = 90° + 30°
🔹 Sum = 120°
🔹 Third angle = 180° − 120°
🔹 Third angle = 60°

✔️ Final Answer: 60°

🔒 ❓ (d) 75°, 45°

📌 ✅ Answer:
🔹 Sum of the two given angles = 75° + 45°
🔹 Sum = 120°
🔹 Third angle = 180° − 120°
🔹 Third angle = 60°

✔️ Final Answer: 60°

🔒 ❓ 2. 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? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

📌 ✅ Answer:
🔹 If two angles are 70°, their sum = 70° + 70°
🔹 Sum = 140°
🔹 Third angle = 180° − 140°
🔹 Third angle = 40°

🔹 Therefore, triangle cannot have all angles 70° because 70° + 70° + 70° = 210° which is greater than 180°.

🔹 If all angles in a triangle are equal, let each angle be x.
🔹 x + x + x = 180°
🔹 3x = 180°
🔹 x = 180° ÷ 3
🔹 x = 60°

✔️ Final Answer:
All angles equal ⇒ each angle must be 60°.

🔒 ❓ 3. Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C?

📌 ✅ Answer:
🔹 Sum of angles in triangle = 180°.
🔹 Given ∠A = 50°.
🔹 Remaining angles sum = 180° − 50°
🔹 Remaining sum = 130°

🔹 Since ∠B = ∠C, let each be x.
🔹 x + x = 130°
🔹 2x = 130°
🔹 x = 130° ÷ 2
🔹 x = 65°

✔️ Final Answer:
∠B = 65°
∠C = 65°

🔵 9. FIGURE IT OUT ?

🔒 ❓ 1. Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.

📌 ✅ Answer:

🔹 Step 1: Draw a line segment BC = 5 cm.

🔹 Step 2: With centre B and radius 6 cm, draw an arc.

🔹 Step 3: With centre C and radius 5 cm, draw another arc cutting the previous arc at point A.

🔹 Step 4: Join A to B and A to C.
Triangle ABC is constructed.

🔹 Step 5: To construct altitude from A to BC:
Place the compass at A and draw an arc cutting BC at two points.

🔹 Step 6: With those two points as centres and equal radius, draw arcs intersecting below BC.

🔹 Step 7: Join A to the point of intersection of these arcs.

The line drawn from A perpendicular to BC is the altitude.

🔒 ❓ 2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.

📌 ✅ Answer:

🔹 Step 1: Draw a line segment RY = 4 cm.

🔹 Step 2: At point R, construct an angle of 140°.

🔹 Step 3: On one arm of the angle, mark point T such that RT = 7 cm.

🔹 Step 4: Join T to Y.
Triangle TRY is constructed.

🔹 Step 5: To construct altitude from T to RY:
With centre T, draw an arc cutting RY at two points.

🔹 Step 6: With those two points as centres and equal radius, draw arcs intersecting below RY.

🔹 Step 7: Join T to the intersection point of arcs.

This line is perpendicular to RY and is the required altitude.

🔒 ❓ 3. Construct a right-angled triangle ΔABC with ∠B = 90°, AC = 5 cm. How many different triangles exist with these measurements?

📌 ✅ Answer:
🔹 Step 1: Draw a line segment AC = 5 cm.

🔹 Step 2: Find the midpoint O of AC.

🔹 Step 3: With centre O and radius OA, draw a circle (so AC is a diameter).

🔹 Step 4: Choose any point B on the circle, on either side of AC (but not at A or C).

🔹 Step 5: Join AB and BC.

🔹 Since AC is a diameter, the angle subtended by AC at any point B on the circle is 90°.

🔹 Therefore, ∠B = 90° for every such choice of B.

🔹 Because point B can be chosen at infinitely many positions on the circle (except A and C), infinitely many different triangles are possible.

✔️ Final Answer: Infinitely many triangles.

🔒 ❓ 4. Through construction, explore if it is possible to construct:
(i) an equilateral triangle that is right-angled
(ii) an equilateral triangle that is obtuse-angled
Also construct an isosceles triangle that is
(i) right-angled
(ii) obtuse-angled.

📌 ✅ Answer:

🔹 In an equilateral triangle, all angles are 60°.
So it cannot be right-angled (90°) or obtuse-angled (> 90°).

Therefore:
(i) Not possible
(ii) Not possible

🔹 An isosceles triangle has two equal sides.

(i) Right-angled isosceles triangle is possible.
Example: angles 45°, 45°, 90°.

(ii) Obtuse-angled isosceles triangle is possible.
Example: angles 100°, 40°, 40°.

Important Notice : The lessons and study materials up to July–August 2025 have been sent. Soon, after receiving feedback from teachers and students, we will deliver the latest and more innovative study materials for the upcoming months.

Important Notice : The lessons and study materials up to July–August 2025 have been sent. Soon, after receiving feedback from teachers and students, we will deliver the latest and more innovative study materials for the upcoming months.

Class 7 : Maths – Lesson 7. A Tale of Three Intersecting Lines

EXPLANATION AND ANALYSIS

🔵 Introduction: When Lines Meet in Different Ways

🟢 Understanding Intersecting Lines Again

🔵 Three Intersecting Lines

🟢 Angles Formed by Three Intersecting Lines

🔵 Adjacent Angles

🟢 Vertically Opposite Angles

🔵 Angle Sum Around a Point

🟡 Special Angle Relationships with Three Lines

🔴 Common Mistakes to Avoid

🟢 Importance of Studying Three Intersecting Lines

📘 Summary

📝 Quick Recap

TEXTBOOK QUESTIONS

OTHER IMPORTANT QUESTIONS

🔵 Section A – Very Short Answer (1 × 6 = 6 marks)

🟢 Section B – Short Answer I (2 × 6 = 12 marks)

🟡 Section C – Short Answer II (3 × 10 = 30 marks)

🔴 Section D – Long Answer (4 × 8 = 32 marks)

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