A Tale of Three Intersecting Lines | CBSE Class 7 Maths Notes

This note covers triangle notation, constructions using sides and angles, triangle inequality, the angle sum property, exterior angles, altitudes and the classification of triangles.

What are the parts of a triangle and how are they named?

A triangle is a closed shape formed by three line segments. A line segment is a straight part of a line with two endpoints. The segments are the triangle's sides; their three meeting points are its vertices. One such corner is called a vertex.

The letters A, B and C can label these vertices. The symbol ∆ means triangle, so ∆ABC means the triangle with vertices A, B and C. The vertices may be written in any order when naming the triangle.

How do side and angle names work?

AB names the segment joining A and B and, in a length statement, its length. The three sides of ∆ABC are AB, BC and CA. A base is the side selected as the starting side for a construction or as the reference for height.

An angle is the opening between two arms meeting at a point. The symbol ∠ means angle. In ∠CAB, the middle letter A identifies the vertex, and AC and AB are its arms. Within ∆ABC, this angle can also be written ∠A.

If the three chosen vertices lie on one straight line, their connecting segments do not enclose a triangular region. This explains why merely marking three points is insufficient: the points must form a closed triangle when joined.

How is an equilateral triangle constructed with a compass?

An equilateral triangle has all three sides equal in length. Constructing one with a marked ruler is possible, but it might require several trials. Making the third vertex the correct distance from one endpoint does not by itself fix its distance from the other endpoint.

A circle consists of points in a flat surface at a fixed distance from a centre. A compass draws circles or parts of circles at that distance. The radius is that distance; an arc is part of a circle. These ideas allow both required distances to be satisfied together.

How do the two arcs locate the third vertex?

Worked example 1. Construct an equilateral triangle whose sides are 4 cm. Here cm means centimetres, a unit of length, and the symbol = means equals.

Answer: Draw AB = 4 cm. Draw an arc of radius 4 cm with centre A and another of radius 4 cm with centre B. Name an intersection C and join AC and BC. Then AB = AC = BC = 4 cm.

An intersection is a point where the arcs meet. Because C lies on the arc centred at A, AC is 4 cm. Because it also lies on the arc centred at B, BC is 4 cm. The base already has this length.

Make the arcs sufficiently long to meet. Complete circles are unnecessary when their relevant arcs locate C. The construction replaces repeated guesses with a point that meets both distance requirements.

The explanation matters as much as the drawing: the equality of the sides follows from the compass settings and the base length, rather than from the triangle merely looking symmetrical.

How can a triangle be constructed when all three sides are given?

The same compass method works when the three side lengths differ. Select one given length as the base. The other two lengths determine distances from the base endpoints to the third vertex. Their arcs must meet to give that vertex.

What are the construction steps?

Worked example 2. Construct ∆ABC with AB = 4 cm, AC = 5 cm and BC = 6 cm.

Answer: Use AB as the base. The following steps locate C by satisfying the 5 cm and 6 cm conditions together.

1. Draw the base AB with length 4 cm.
2. With A as centre, draw a sufficiently long arc of radius 5 cm.
3. With B as centre, draw an arc of radius 6 cm that intersects the first arc.
4. Name the intersection C. Join AC and BC to complete ∆ABC.

Every point on the first circle is 5 cm from A. Every point on the second circle is 6 cm from B. A shared point therefore meets both conditions. This is why joining an intersection to A and B produces the required side lengths.

The sequence shows a 4 cm base, a circle centred at A with radius 5 cm, and an intersecting circle centred at B with radius 6 cm.

See Figs. 7.1, 7.2 and 7.3 in your NCERT textbook

When full circles meet at two points, either intersection can be used. The essential requirement is that the chosen point belongs to both circles. A point selected on just one circle satisfies only one of the two distance conditions.

An isosceles triangle has two sides of equal length. The construction method also applies to the given sets 4 cm, 4 cm, 6 cm and 1 cm, 5 cm, 5 cm.

Why do some sets of three lengths fail to form a triangle?

The triangle inequality requires each side length to be smaller than the sum of the other two. The symbol < means less than; > means greater than. The symbol + means addition. The comparison is strict: equality does not satisfy it.

Property: Each side is shorter than the other two together

For ∆ABC, the three comparisons are AB < AC + CB, BC < BA + AC and CA < CB + BA. Reversing a segment's endpoint letters does not change its length.

Think of a tent, a tree and a pole at the three corners of a triangular arrangement. The direct straight path between two corners is shorter than going through the third. Each triangle side must therefore be shorter than the route along the other two sides.

Worked example 3. Could a triangle have BC = 10 cm, AB = 15 cm and CA = 30 cm?

Answer: No. Although 10 cm is less than 45 cm and 15 cm is less than 40 cm, the remaining comparison fails: 30 cm is greater than 10 cm + 15 cm = 25 cm.

Note: A rough sketch carrying three length labels does not establish that the triangle exists. The lengths must satisfy the required comparisons.

Changing the positions of these three lengths cannot remove the difficulty. The 30 cm side would still exceed the sum of the 10 cm and 15 cm sides. Two successful comparisons therefore cannot compensate for a failed third comparison.

How do circles explain the triangle inequality?

To examine whether a triangle exists, choose the longest length as base AB. Use the two smaller lengths as the radii of circles centred at A and B. The radii are the two distances that the third vertex must have from the base endpoints.

What happens when the circles touch, separate or cross?

In the touching case, the common point lies on the base, so it does not create a triangular region. In the separated case, no common point satisfies both distance requirements. When the circles cross, either crossing supplies a third vertex away from the base.

Worked example 4. Determine whether the lengths 4 cm, 5 cm and 8 cm form a triangle.

Answer: Take AB = 8 cm and draw circles of radii 4 cm and 5 cm centred at A and B. Since 8 < 4 + 5 = 9, the circles intersect at two points. Either intersection completes the required triangle.

Two circles with centres A and B overlap and meet above and below AB. Their radii are marked 4 cm and 5 cm, and the upper meeting point is joined to the centres.

See Fig. 7.5 in your NCERT textbook

Before analysing the circles, the successful length comparisons alone leave the tentative conclusion that a triangle may or may not exist. The circle argument supplies the missing justification: satisfying the triangle inequality produces the necessary intersections.

The resulting conclusion works in both directions. Triangle sides satisfy the inequality, and a set of three lengths satisfying it can be used to construct a triangle.

How can possible side lengths be checked efficiently?

For three positive lengths, meaning lengths greater than zero, it is enough to compare the longest length with the sum of the other two. Each smaller length is already less than the longest length plus another positive length. This accounts for the other comparisons without separately calculating them.

What is the checking procedure?

1. Make sure the three lengths are expressed in the same unit.
2. Identify a longest length in the set.
3. Add the other two lengths.
4. A triangle exists if this sum is greater than the longest length; equality or a smaller sum rules it out.

Worked example 5. Can lengths 3, 6 and 9 form a triangle? All three lengths use the same unit.

Answer: No. The longest length is 9, and the other two add to 3 + 6 = 9. Equality gives circles touching at one point rather than a triangle.

Worked example 6. Can lengths 1, 100 and 100 form a triangle? All three lengths use the same unit.

Answer: Yes. A longest length is 100, while the other two add to 1 + 100 = 101. Since 100 < 101, the triangle inequality is satisfied.

With two fixed lengths, there can be many choices for the third length, including decimal values. For lengths 1 and 100, the third length can be any number strictly between 99 and 101, with all lengths in the same unit.

The endpoints are excluded. At either endpoint one length equals the sum of the other two, so no triangular region forms. The size difference between two sides alone is therefore not a valid reason for rejecting a triangle.

How is a triangle constructed from two sides and their included angle?

The included angle is the angle between the two given sides. For sides AB and AC, it is ∠A because the sides meet at A. The symbol ° means degrees, the unit used here for measuring angles.

Knowing the included angle fixes how the two given sides open away from their common vertex. After drawing one side, draw the other arm at the specified angle. Then mark the required second length along that arm and connect the remaining endpoints.

Which measurement goes at each step?

Worked example 7. Construct ∆ABC with AB = 5 cm, AC = 4 cm and ∠A = 45°.

Answer: Use the 5 cm side as the base and place the 45° angle at A, the common endpoint of the two given sides.

1. Draw AB = 5 cm.
2. At A, draw the other angle arm so that it makes 45° with AB.
3. Mark C on this arm at a distance of 4 cm from A.
4. Join BC to complete the triangle.

The position of the angle is part of the information. Placing 45° at B would change the construction because the given 4 cm and 5 cm sides meet at A. Similarly, the 4 cm distance must be measured from A along the new arm.

Check the finished drawing against all three conditions: the base length, the second given side length and the angle between them. No value for BC is supplied, so it is obtained by joining the constructed points rather than assigned beforehand.

When do two angles and their included side form a triangle?

The included side is the side shared by the two given angles. In ∆ABC, AB is the included side between ∠A and ∠B. Draw this side first, then construct the given angles at its endpoints on the same side of the base.

Worked example 8. Construct ∆ABC with AB = 5 cm, ∠A = 45° and ∠B = 80°.

Answer: Draw AB = 5 cm. At A draw the 45° angle arm, and at B draw the 80° angle arm on the same side of AB. Their intersection is C. The segments AC and BC complete the triangle.

Property: Two triangle angles have a sum below 180°

A straight angle measures 180°, and a right angle measures 90°. Two given positive angles can belong to a triangle when their sum is less than 180°. If their sum is greater than or equal to 180°, they cannot form two angles of a triangle.

To understand the limiting case, take ∠A = 40°. The line through B parallel to the other arm at A gives ∠B = 140°. Parallel lines are lines in the same plane, or flat surface, that do not meet. The base crosses these parallel lines.

A transversal is a line crossing two lines. The interior angles on its same side lie between those lines, on the same side of the crossing line. For parallel lines, these angles add to 180°, giving 40° + 140° = 180°.

With the 40° angle fixed, a second angle greater than or equal to 140° does not produce a triangle. Changing the base length does not alter this condition.

Worked example 9. Can 70° and 30° be two triangle angles? Can 35° and 150°?

Answer: The first pair works because 70° + 30° = 100°, less than 180°. The second pair fails because 35° + 150° = 185°, greater than 180°.

Why do the three angles of a triangle add to 180°?

The angle sum property gives an exact relationship between the three angles. An extra line makes the relationship visible: draw a line through one vertex parallel to the opposite side, meaning the side that does not meet that vertex.

Property: The angle sum of a triangle is 180°

Consider ∆ABC and draw line XY through A parallel to BC, with X and Y on opposite sides of A. The sides AB and AC act as transversals. They allow the angles at B and C to be compared with angles at A.

Alternate interior angles lie between the two lines on opposite sides of a transversal. When the two lines are parallel, these angles are equal. Here the relevant pairs are ∠ABC and ∠XAB, and ∠BCA and ∠YAC.

1. Draw XY through A parallel to BC.
2. Using AB as the transversal, identify ∠XAB = ∠ABC.
3. Using AC as the transversal, identify ∠YAC = ∠BCA.
4. The three angles ∠XAB, ∠BAC and ∠YAC together form the straight angle at A. Replace the outside two by their equal triangle angles.

Therefore, ∠A + ∠B + ∠C = 180°. The argument uses the relationships created by parallel lines, so it applies to any triangle, rather than depending on the apparent size or shape of a particular drawing.

Triangle ABC has BC as its lower side. Line XY passes through A above BC. Matching coloured angles at B and C also appear beside the angle at A, together filling a straight angle.

See Fig. 7.7 in your NCERT textbook

Folding a triangular paper cut-out offers a way of verifying the angle sum property. The proof explains why the sum is 180°, while the folding activity allows the three angles to be brought together and inspected.

When two angles are fixed and the included side is changed, measurements might show that the side length has no effect (or a very small effect) on the third angle. The exact angle sum property then determines that third angle without construction or measurement.

How are missing angles and exterior angles calculated?

Once two angles are known, subtract their sum from 180° to find the third. The symbol − means subtraction. Keep the angle names attached to the calculation so that the answer identifies the angle being found.

How is an unknown interior angle found?

Worked example 10. In ∆ABC, ∠B = 50° and ∠C = 70°. Find ∠A.

Answer: The angle sum gives ∠A + 50° + 70° = 180°. Thus ∠A = 180° − 120° = 60°.

Worked example 11. In ∆ABC, ∠B = ∠C and ∠A = 50°. Find both equal angles.

Answer: Together ∠B and ∠C measure 180° − 50° = 130°. Since they are equal, each is half of 130°. Therefore ∠B = ∠C = 65°.

The equality condition in the second example is essential. Knowing only the 50° angle fixes the sum of the other two angles; it does not separately fix each one. Both the total and the equality are needed to find 65°.

What is an exterior angle?

An exterior angle is formed between an extended side of a triangle and the other side at that vertex. Extend BC beyond C to a point D. The angle between CA and CD is ∠ACD, an exterior angle at C.

Worked example 12. In ∆ABC, ∠A = 50° and ∠B = 60°. BC is extended beyond C to D. Find ∠ACD.

Answer: First, ∠ACB = 180° − 50° − 60° = 70°. The angles ∠ACB and ∠ACD form a straight angle. Therefore ∠ACD = 180° − 70° = 110°.

The interior angle at C occurs in two totals: the three triangle angles sum to 180°, and the interior and exterior angles at C also sum to 180°. Subtracting the common interior angle gives ∠ACD = ∠A + ∠B.

This compares the exterior angle with the two triangle angles away from C. It does not say that the exterior angle equals the interior angle next to it. Their relationship is instead that they complete a straight angle.

What is an altitude and how is it constructed?

Perpendicular lines meet at a right angle. An altitude of a triangle is a perpendicular segment drawn from a vertex to its opposite side or to that side extended. Its length measures the vertex's height above the line containing that side.

In ∆ABC, take BC as the base and let D be the point where the perpendicular from A meets its line. Then AD is an altitude. The height of a triangle generally refers to the length of the altitude to whichever side is taken as the base.

Triangle ABC has a segment AD drawn down to BC. D lies between B and C, and a right-angle mark shows that AD meets BC perpendicularly.

See Fig. 7.8 in your NCERT textbook

How do a ruler and set square help?

A set square is a drawing tool with a right-angled corner. Using it with a ruler gives a more precise 90° angle than trying to construct the altitude with only a ruler.

1. Align the ruler with base BC.
2. Place one edge of the set square's right angle against the ruler.
3. Slide the set square along the ruler until its perpendicular edge passes through A.
4. Draw along that edge from A to the line containing BC to obtain the altitude.

The meeting point need not lie within the side segment. If necessary, extend the base and drop the perpendicular to the extended line. Restricting the construction to the inside of the triangle would miss such altitudes.

A side can itself be an altitude. In a triangle with a right angle at B, AB is perpendicular to BC, so AB is the altitude from A to BC. A triangle with one right angle is called a right-angled triangle, or right triangle.

There are also altitudes from B to the line containing AC and from C to the line containing AB. Each is identified by its starting vertex and the opposite side used for its perpendicular measurement.

How are triangles classified by their sides and angles?

Triangle classification can use side lengths or angle measures. The first asks which sides have equal lengths. The second asks whether the angles are acute, right or obtuse. These are different ways of examining a triangle, so state the basis of a classification.

What are the types based on sides?

A scalene triangle therefore differs from the equal-sided constructions. In checking a side classification, use the given lengths or the construction conditions. The way a triangle happens to be drawn is not a substitute for information about its sides.

What are the types based on angles?

An acute angle is greater than 0° and less than 90°. An obtuse angle is greater than 90° and less than 180°. These definitions, together with the 90° right angle, give three angle-based triangle types.

One acute angle is insufficient to identify an acute-angled triangle. All three must be checked. The angle sum property explains why two right angles, or two obtuse angles, cannot occur in a triangle: they leave no positive measure for the third angle.

Construction exercises can combine these ideas with altitudes. For example, a triangle with BC = 5 cm, AB = 6 cm and CA = 5 cm is isosceles. After constructing it from the three side lengths, an altitude can be drawn from A to BC.

Glossary

• Triangle — A closed shape formed by three line segments joining three vertices.
• Vertex — A corner point where two sides of a triangle meet.
• Base — A chosen triangle side used for construction or as the reference for height.
• Radius — The fixed distance from a circle's centre to any point on that circle.
• Arc — A part of a circle used to locate points at a fixed distance.
• Equilateral triangle — A triangle in which all three sides have the same length.
• Isosceles triangle — A triangle with two sides of equal length among its three sides.
• Scalene triangle — A triangle whose three sides all have different lengths.
• Triangle inequality — The requirement that each side length be less than the sum of the other two.
• Included angle — The angle formed between the two given sides at their common vertex.
• Included side — The triangle side shared by the two given angles at its endpoints.
• Angle sum property — The result that the three interior angles of any triangle total 180°.
• Exterior angle — An angle between an extended triangle side and the other side at that vertex.
• Altitude — A perpendicular segment from a triangle vertex to its opposite side or that side extended.
• Acute-angled triangle — A triangle in which each of the three angles is acute.

Common errors and misconceptions

• Misconception: Any three positive lengths form a triangle. Correct: They must satisfy the triangle inequality. The longest length must be smaller than the sum of the other two.
• Misconception: Equality is allowed in the triangle inequality. Correct: When two lengths add to the third, the construction circles touch on the base and no triangular region forms.
• Misconception: Two successful side comparisons are enough. Correct: Every comparison must hold. With lengths 10 cm, 15 cm and 30 cm, the longest-side comparison fails.
• Misconception: The included angle can be placed at either end of a given side. Correct: It belongs at the vertex where the two specified sides meet.
• Misconception: Increasing the base can make a triangle from two angles whose sum is 180°. Correct: Their sum must be less than 180°, regardless of the base length.
• Misconception: An exterior angle equals its neighbouring interior angle. Correct: The two together form 180°. The exterior angle equals the sum of the other two interior angles.
• Misconception: Every altitude lies inside the triangle. Correct: An altitude may meet an extended side; in a right triangle, a side can itself be an altitude.
• Misconception: A triangle with one acute angle is acute-angled. Correct: All three angles must be acute for this classification.

Exam-style questions with model answers

Q1. Can lengths 3, 6 and 9, measured in the same unit, form a triangle? Give a reason. [2 marks]

1. The longest length is 9, while the other two add to 3 + 6 = 9.
2. No triangle exists because the triangle inequality requires the sum of the two smaller lengths to be greater than the longest.

Q2. In ∆ABC, ∠B = 50° and ∠C = 70°. Find ∠A, stating the property used. [2 marks]

1. The angle sum property gives ∠A + ∠B + ∠C = 180°, so ∠A + 50° + 70° = 180°.
2. The two known angles total 120°. Therefore the missing angle is ∠A = 180° − 120° = 60°.

Q3. Describe how to construct ∆ABC with AB = 5 cm, AC = 4 cm and the included angle ∠A = 45°. [4 marks]

1. Draw the given side AB with length 5 cm. Use this segment as the base for the construction.
2. At A, draw a second arm making 45° with AB. A is the common endpoint of the two given sides.
3. On this new arm, mark C so that its distance from A is 4 cm, giving AC = 4 cm.
4. Join B to C. The triangle now has the specified base, second side and angle included between them.

Q4. In ∆ABC, ∠A = 50° and ∠B = 60°. Side BC is extended beyond C to D. Find the exterior angle ∠ACD, showing the steps. [3 marks]

1. The interior angles of the triangle total 180°. The two given angles add to 50° + 60° = 110°.
2. Therefore the interior angle at C is ∠ACB = 180° − 110° = 70°. This is the angle between CA and CB.
3. Since CB and CD are opposite arms of a straight line, ∠ACB + ∠ACD = 180°. Hence ∠ACD = 180° − 70° = 110°.

Q5. Prove that the three angles of any triangle ABC sum to 180°, using a line through A parallel to BC. [5 marks]

1. Draw line XY through A parallel to BC, with X and Y on opposite sides of A. This produces new angles beside ∠BAC.
2. Using AB as a transversal of the parallel lines, identify the equal alternate interior angles: ∠XAB = ∠ABC.
3. Using AC as a transversal of the same parallel lines, identify the second equal pair: ∠YAC = ∠BCA.
4. The angles ∠XAB, ∠BAC and ∠YAC together make the straight angle along XY. Their sum is therefore 180°.
5. Replace ∠XAB and ∠YAC by their equal angles at B and C. This gives ∠ABC + ∠BAC + ∠BCA = 180°, proving the angle sum property.

Q6. Construct ∆ABC with AB = 4 cm, AC = 5 cm and BC = 6 cm using a ruler and compass. Explain why the construction satisfies the given lengths. [5 marks]

1. Draw AB = 4 cm with the ruler. This fixes the base and the two centres from which the remaining distances will be measured.
2. With A as centre, draw a sufficiently long arc of radius 5 cm. The required point C must lie on this arc.
3. With B as centre, draw an arc of radius 6 cm that intersects the first arc. Choose a meeting point and label it C.
4. Join AC and BC to complete the triangle. Because C lies on the first arc, AC equals its radius, 5 cm.
5. Because C also lies on the second arc, BC equals its radius, 6 cm. Along with the 4 cm base, all three specified lengths are satisfied.

Q7. In a triangle ABC, the perpendicular from A meets BC at D. Explain how to construct AD using a ruler and set square. [3 marks]

1. Keep the ruler aligned with BC. Place one edge of the right angle of the set square against this ruler.
2. Slide the set square along the ruler until its other right-angle edge passes through vertex A. Keep the ruler aligned with BC.
3. Draw along that edge from A to BC and label the meeting point D. AD meets BC at 90°, so it is the required altitude.

Q8. In ∆ABC, ∠A = 50° and ∠B = ∠C. Find ∠B and ∠C, explaining how both conditions are used. [3 marks]

1. Use the angle sum property to write ∠B + ∠C = 180° − 50°. Thus the two unknown angles together measure 130°.
2. The question also gives ∠B = ∠C. This means the 130° total must be shared equally between these two angles.
3. Half of 130° is 65°, so ∠B = 65° and ∠C = 65°. Both the angle sum and the equality condition have been used.

Key takeaways

• A triangle has three sides, three vertices and three angles; its name uses the letters assigned to its vertices.
• Intersecting compass arcs locate a third vertex by satisfying its required distances from both endpoints of the base.
• Triangle inequality requires each side to be shorter than the sum of the other two; equality does not work.
• For three positive lengths, checking the longest against the sum of the other two is enough to decide triangle existence.
• Two sides with their included angle, or two suitable angles with their included side, allow triangle construction.
• The angles of any triangle total 180°, and two given triangle angles must have a sum below 180°.
• An exterior angle and the interior angle beside it form a straight angle; the exterior angle equals the other two interior angles together.
• An altitude is perpendicular to the chosen base or its extension; its length gives the height relative to that base.
• Side classification uses equalities of length, while angle classification distinguishes acute-angled, right-angled and obtuse-angled triangles.

Test yourself

Why is ∠CAB an angle at A?

The middle letter names the vertex. The angle's two arms are AC and AB, which meet at A.

Why does the compass construction of an equilateral triangle with side 4 cm work?

The arc intersection lies 4 cm from each endpoint of the 4 cm base, making all three sides equal.

Can lengths 1, 100 and 100 in the same unit form a triangle?

Yes. A longest length is 100, and the other two total 101. The required strict inequality is satisfied.

With side lengths 1 and 100, what values can the third length take in the same unit?

Any value strictly between 99 and 101 is possible. The endpoints are excluded because equality does not satisfy the triangle inequality.

Can 35° and 150° be two angles of a triangle?

No. They add to 185°, exceeding 180°, so they cannot be two angles of a triangle.

If two triangle angles are 36° and 72°, what is the third?

The third angle is 180° − 36° − 72° = 72°, using the angle sum property.

If ∠B = 90° in ∆ABC, why can AB be an altitude?

AB is perpendicular to BC. It is therefore the altitude from A when BC is taken as the base.

Does one acute angle make a triangle acute-angled?

No. All three angles must be acute before the triangle can be classified as acute-angled.