A Tale of Three Intersecting Lines

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CLASS VII Mathematics ~6 marks/year Ch 7 of 15
A Tale of Three Intersecting Lines

Class 7 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Three line segments meeting pairwise can enclose a triangle — the simplest closed figure, with three sides, three angles and three vertices.
  • The three angles of any triangle always add up to $180°$ — the angle sum property.
  • The sum of any two sides of a triangle is greater than the third side — the triangle inequality; this decides whether three lengths can even form a triangle.
  • An exterior angle of a triangle equals the sum of the two opposite interior angles.
  • Triangles are classified by sides (scalene, isosceles, equilateral) and by angles (acute, right, obtuse).
  • You learn to construct triangles from given measurements and to reason about which sets of measurements actually produce a triangle.
  • Weightage: ~6 marks/year — angle-finding problems, triangle-inequality checks, and one construction/reasoning question.
Detailed Notes

1. When three lines make a triangle

Take three straight lines in a plane. If no two are parallel and they do not all pass through one point, then each pair crosses at a point, giving three crossing points. Joining these three points with the segments between them encloses a triangle. So a triangle is, quite literally, the tale of three intersecting lines.

Triangle: a closed figure formed by three line segments, having three sides, three vertices (corner points) and three angles. A triangle with vertices $A, B, C$ is written $\triangle ABC$.

If two of the three lines are parallel, they never meet, so only two crossing points exist and no triangle is formed. If all three lines pass through a single point, again no enclosed region appears. So the triangle needs three lines in "general position".

2. The angle sum property

The most important fact about a triangle is that its three interior angles always add to a straight angle.

Angle sum property: In any triangle, $\angle A + \angle B + \angle C = 180°$.
Worked example — why the angles add to 180°

Through one vertex of the triangle draw a line parallel to the opposite side. The two base angles of the triangle become alternate interior angles with the angles formed at the vertex along the parallel line. These three angles at the vertex together make a straight line $= 180°$, and they equal the three angles of the triangle. Hence the angle sum is $180°$.

Worked example — finding the third angle

Two angles of a triangle are $65°$ and $48°$. The third is $180° - 65° - 48° = 67°$.

A direct consequence: a triangle can have at most one right angle and at most one obtuse angle, because two such angles would already use up $180°$ or more.

3. The exterior angle property

If you extend one side of a triangle beyond a vertex, the angle formed outside, between the extension and the adjacent side, is an exterior angle. It has a beautiful relationship with the two angles at the far corners.

Exterior angle property: an exterior angle of a triangle equals the sum of the two interior opposite angles. So if the exterior angle at $C$ is $\angle ACD$, then $\angle ACD = \angle A + \angle B$.
Worked example — using the exterior angle

In $\triangle ABC$, $\angle A = 50°$ and $\angle B = 60°$. The exterior angle at $C$ is $50° + 60° = 110°$. Check: interior $\angle C = 180° - 50° - 60° = 70°$, and indeed $70° + 110° = 180°$ (linear pair), confirming the result.

This property is just the angle sum property in disguise: the exterior angle plus the adjacent interior angle is $180°$ (linear pair), and the three interior angles also add to $180°$, so the exterior angle must equal the other two interiors.

4. The triangle inequality

Not every three lengths can form a triangle. To close up into a triangle, the two shorter sides together must be able to "reach across" the longest side.

Triangle inequality: the sum of the lengths of any two sides of a triangle is greater than the length of the third side. $a + b > c$, $b + c > a$, and $c + a > b$.
Worked example — can these form a triangle?

Lengths $3$ cm, $4$ cm, $8$ cm. Check the largest: is $3 + 4 > 8$? No, $7 < 8$. So these cannot form a triangle. Lengths $5, 6, 9$: check $5 + 6 = 11 > 9$, $6 + 9 > 5$, $5 + 9 > 6$ — all true, so they can form a triangle.

Shortcut: you only need to check that the sum of the two smaller sides exceeds the largest side; if that holds, the other two checks hold automatically.

A related fact: the difference of any two sides is less than the third side. So each side lies strictly between the difference and the sum of the other two.

5. Classifying triangles by sides

Triangles are sorted by how many sides are equal.

Type Sides Angles
Scaleneall three sides differentall three angles different
Isoscelesexactly two sides equalthe two base angles equal
Equilateralall three sides equalall three angles $= 60°$

An important link: equal sides face equal angles. In an isosceles triangle, the angles opposite the two equal sides are equal. In an equilateral triangle, since all sides are equal, all angles are equal, and since they sum to $180°$, each is $180° \div 3 = 60°$.

6. Classifying triangles by angles

Type Largest angle
Acute-angledall three angles less than $90°$
Right-angledone angle exactly $90°$
Obtuse-angledone angle greater than $90°$
Worked example — name the triangle

A triangle has angles $90°, 45°, 45°$. It has a right angle, so it is right-angled; two angles ($45°$) are equal, so two sides are equal, making it also isosceles. We call it a right isosceles triangle.

7. Constructing triangles

To draw a triangle accurately you need exactly enough information. Common cases (using ruler, compasses and protractor):

  • SSS — three sides given. Draw the base, then strike arcs of the other two lengths from the two endpoints; their crossing is the third vertex.
  • SAS — two sides and the included angle. Draw one side, set the given angle at one end, mark the second side along it, and join.
  • ASA — two angles and the included side. Draw the side, draw the two angles at its ends, and extend the arms until they meet.
Construction — SSS triangle with sides 4 cm, 5 cm, 6 cm

1. Draw the base $BC = 6$ cm. 2. With centre $B$ and radius $5$ cm, draw an arc. 3. With centre $C$ and radius $4$ cm, draw another arc cutting the first at $A$. 4. Join $AB$ and $AC$. $\triangle ABC$ is the required triangle. The construction works only because $4 + 5 > 6$ (triangle inequality), so the arcs actually meet.

Notice: if the triangle inequality fails, the two arcs never intersect, and no triangle can be drawn — geometry and arithmetic agree.

8. Common mistakes to avoid

  • Forgetting to check the triangle inequality before assuming three lengths form a triangle.
  • Using the exterior angle as equal to all three interior angles — it equals only the two opposite interiors.
  • Thinking a triangle can have two right angles or two obtuse angles.
  • Mixing up "equal sides face equal angles" — the equal angle is opposite the equal side, not next to it.
  • Calling every triangle with two equal angles "equilateral"; it is isosceles unless all three are equal.

9. Quick revision checklist

  • Three intersecting lines in general position form a triangle.
  • Angle sum $= 180°$; exterior angle $=$ sum of two opposite interiors.
  • Triangle inequality: sum of two sides $>$ third side; difference $<$ third side.
  • By sides: scalene, isosceles, equilateral; equal sides face equal angles.
  • By angles: acute, right, obtuse; at most one right or obtuse angle.
  • Construct with SSS, SAS, ASA when enough measurements are given.
Practice MCQs
1. The sum of the three interior angles of a triangle is:
  1. $90°$
  2. $180°$
  3. $270°$
  4. $360°$
Answer: (B) The angle sum property gives $180°$.
2. Two angles of a triangle are $70°$ and $60°$. The third angle is:
  1. $40°$
  2. $50°$
  3. $60°$
  4. $70°$
Answer: (B) $180° - 70° - 60° = 50°$.
3. Which set of lengths can form a triangle?
  1. $2, 3, 6$
  2. $4, 5, 9$
  3. $5, 6, 10$
  4. $1, 2, 4$
Answer: (C) $5 + 6 = 11 > 10$; the others fail since their two smaller sides do not exceed the largest.
4. A triangle with all three sides equal is called:
  1. scalene
  2. isosceles
  3. equilateral
  4. right-angled
Answer: (C) Equal sides on all three make it equilateral.
5. Each angle of an equilateral triangle measures:
  1. $45°$
  2. $60°$
  3. $90°$
  4. $120°$
Answer: (B) $180° \div 3 = 60°$.
6. An exterior angle of a triangle equals:
  1. the adjacent interior angle
  2. the sum of the two opposite interior angles
  3. $180°$
  4. the largest interior angle
Answer: (B) Exterior angle $=$ sum of the two interior opposite angles.
7. In a triangle, $\angle A = 40°$ and $\angle B = 75°$. The exterior angle at $C$ is:
  1. $65°$
  2. $105°$
  3. $115°$
  4. $120°$
Answer: (C) $40° + 75° = 115°$.
8. A triangle with one angle of $110°$ is:
  1. acute-angled
  2. right-angled
  3. obtuse-angled
  4. equilateral
Answer: (C) An angle greater than $90°$ makes it obtuse-angled.
9. How many right angles can a triangle have at most?
  1. $0$
  2. $1$
  3. $2$
  4. $3$
Answer: (B) Two right angles would already total $180°$, leaving nothing for the third.
10. In an isosceles triangle, the angles opposite the equal sides are:
  1. equal
  2. supplementary
  3. complementary
  4. both right angles
Answer: (A) Equal sides face equal angles.
11. The difference of any two sides of a triangle is always:
  1. greater than the third side
  2. equal to the third side
  3. less than the third side
  4. zero
Answer: (C) Each side lies between the difference and the sum of the other two.
12. A triangle has angles $90°, 30°, 60°$. It is:
  1. acute scalene
  2. right scalene
  3. right isosceles
  4. obtuse
Answer: (B) One angle is $90°$ (right) and all angles differ (scalene).
13. To construct a triangle by SSS we are given:
  1. three angles
  2. three sides
  3. two sides and an angle
  4. two angles and a side
Answer: (B) SSS means all three Sides are given.
14. If one angle of an isosceles triangle is $40°$ and it is one of the two equal base angles, the third angle is:
  1. $40°$
  2. $70°$
  3. $100°$
  4. $140°$
Answer: (C) Base angles $40°$ each, so apex $= 180° - 40° - 40° = 100°$.
15. Three lines form a triangle only if:
  1. two of them are parallel
  2. all three pass through one point
  3. no two are parallel and they are not concurrent
  4. they are all perpendicular
Answer: (C) They must cross pairwise at three different points to enclose a region.
Important Questions
Q1. State the angle sum property of a triangle and use it to find the third angle when two angles are $55°$ and $85°$. (2 marks)
Answer: The angle sum property states that the three interior angles of a triangle add up to $180°$. Third angle $= 180° - 55° - 85° = 40°$.
Q2. State and verify the exterior angle property for a triangle with $\angle A = 35°$, $\angle B = 65°$. (3 marks)
Answer: The exterior angle equals the sum of the two interior opposite angles. Exterior angle at $C = 35° + 65° = 100°$. Interior $\angle C = 180° - 35° - 65° = 80°$, and $80° + 100° = 180°$ (linear pair), so the property checks out.
Q3. Can a triangle be formed with sides $6$ cm, $7$ cm and $13$ cm? Justify using the triangle inequality. (2 marks)
Answer: Check the two smaller sides against the largest: $6 + 7 = 13$, which is not greater than $13$. Since the sum is not greater than the third side, the triangle inequality fails, so no triangle can be formed.
Q4. Classify triangles by their sides and by their angles, naming each type. (3 marks)
Answer: By sides: scalene (all sides different), isosceles (two sides equal), equilateral (all three sides equal). By angles: acute-angled (all angles less than $90°$), right-angled (one angle $90°$), obtuse-angled (one angle more than $90°$).
Q5. Explain why an equilateral triangle has each angle equal to $60°$. (2 marks)
Answer: All three sides are equal, and equal sides face equal angles, so all three angles are equal. Since they add to $180°$, each angle $= 180° \div 3 = 60°$.
Q6. Describe the steps to construct a triangle with sides $5$ cm, $6$ cm and $7$ cm (SSS). (3 marks)
Answer: Draw base $BC = 7$ cm. With centre $B$ and radius $6$ cm draw an arc; with centre $C$ and radius $5$ cm draw another arc cutting the first at $A$. Join $AB$ and $AC$. $\triangle ABC$ is the required triangle. The arcs meet because $5 + 6 > 7$.
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More Class 7 Mathematics chapters
Large Numbers Around Us · Arithmetic Expressions · A Peek Beyond the Point · Expressions using Letter-Numbers · Parallel and Intersecting Lines · Number Play · Working with Fractions · Geometric Twins
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