A Tale of Three Intersecting Lines

CHAPTER 7: A TALE OF THREE INTERSECTING LINES

COMPREHENSIVE CLASS 7 MATHEMATICS NOTES

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INTRODUCTION TO TRIANGLES

**A triangle** is the most basic closed shape in geometry. It consists of three essential parts:

• **Three vertices (corner points)**: These are the three corners where sides meet. We label them with capital letters like A, B, C.

• **Three sides (line segments)**: These are the three straight lines connecting pairs of vertices.

• **Three angles**: These are formed where the sides meet at the vertices.

Naming Triangles

Triangles are denoted using the symbol **△** (delta). When naming a triangle, we write the vertices in any order. For example:

Triangle ABC can also be written as △ABC, △BAC, △CAB, etc.

The three angles of a triangle are named using the vertex:

In △ABC, the angles are ∠CAB, ∠ABC, and ∠BCA

We can simply write these as ∠A, ∠B, and ∠C

Important Question

**What happens when three vertices lie on a straight line?** Answer: They do not form a triangle. A triangle requires that the three vertices are not collinear (not on the same line).

Real-Life Indian Context

In Indian architecture, triangular roof structures are common in temples and houses. The stability of these structures depends on the three sides meeting properly at vertices.

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7.1 EQUILATERAL TRIANGLES

**Definition**: An **equilateral triangle** is a triangle in which all three sides have equal lengths. Among all triangles, equilateral triangles are the most symmetric ones.

Properties of Equilateral Triangles

• All three sides are of equal length

• All three angles are equal (each measuring 60°)

• It has three lines of symmetry

• It is the most symmetric triangle possible

Construction of an Equilateral Triangle with Side 4 cm

**Why we need a compass and ruler**: While we could use just a marked ruler, it would require many trials to find the correct position of the third vertex. Using a compass makes the construction efficient and accurate.

**Step-by-Step Construction Method:**

**Step 1**: Draw the base AB = 4 cm using a ruler and pencil.

**Step 2**: Open the compass to a width of 4 cm. Place the compass point on A and draw an arc above the base AB. This arc contains all points that are exactly 4 cm away from A.

**Step 3**: Without changing the compass width, place the compass point on B and draw another arc that intersects the first arc. Let C be the point where the two arcs intersect.

**Step 4**: Join AC and BC using a ruler. The triangle ABC is your required equilateral triangle.

**Why this works**:

The arc from A contains all points 4 cm from A. Point C lies on this arc, so AC = 4 cm.

The arc from B contains all points 4 cm from B. Point C lies on this arc, so BC = 4 cm.

We already drew AB = 4 cm.

Therefore, AB = AC = BC = 4 cm, making it equilateral.

Real-Life Example

In Indian tile patterns and rangoli designs, equilateral triangles are often used because they create perfectly symmetric patterns when arranged together.

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7.2 CONSTRUCTING TRIANGLES WHEN ALL THREE SIDES ARE GIVEN

The Problem: When Sides Are Different Lengths

When a triangle has sides of different lengths (not equilateral), we must use a method to ensure all three conditions are satisfied simultaneously.

**Construct a triangle with sides 4 cm, 5 cm, and 6 cm**

Step-by-Step Construction Method

**Step 1**: Choose one side length to be the base. Let's choose 4 cm. Draw base AB = 4 cm using a ruler.

Point A is one endpoint

Point B is the other endpoint

Length = 4 cm

**Step 2**: We need to find point C such that:

AC = 5 cm (second given side)

BC = 6 cm (third given side)

**Step 3**: Open the compass to 5 cm. Place the compass point on A and draw a long arc. This arc contains all points exactly 5 cm away from A. Point C must lie somewhere on this arc.

**Step 4**: Without changing the compass setting, place the compass point on B. We need to change the compass setting to 6 cm. Place the compass point on B and draw another arc. This arc contains all points exactly 6 cm away from B. Point C must also lie on this arc.

**Step 5**: The point where both arcs intersect is C (the third vertex). Mark this point clearly.

**Step 6**: Join AC and BC using a ruler. Triangle ABC is complete.

Why This Construction Works

• Point C lies on the arc centered at A with radius 5 cm, so AC = 5 cm ✓

• Point C lies on the arc centered at B with radius 6 cm, so BC = 6 cm ✓

• We constructed AB = 4 cm ✓

All three conditions are satisfied.

Understanding Different Types of Triangles

**Equilateral Triangle**: All three sides are equal

Example: 5 cm, 5 cm, 5 cm

**Isosceles Triangle**: Exactly two sides are equal

Example: 4 cm, 4 cm, 6 cm

Example: 5 cm, 5 cm, 8 cm

**Scalene Triangle**: All three sides have different lengths

Example: 4 cm, 5 cm, 6 cm

Example: 3 cm, 4 cm, 5 cm

Practice Constructions

Construct triangles with the following side lengths (in cm):

(a) 4, 4, 6 (isosceles)

(b) 3, 4, 5 (scalene)

(c) 1, 5, 5 (isosceles)

(d) 4, 6, 8 (scalene)

(e) 3.5, 3.5, 3.5 (equilateral)

Real-Life Example

In Indian engineering and bridge construction, triangular frames are used because once three sides are fixed, the triangle cannot change shape. This stability is critical for structures.

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ARE TRIANGLES POSSIBLE FOR ANY LENGTHS?

A crucial question: Can we always construct a triangle for any three given lengths?

Exploration Activity

**Attempt 1**: Try to construct a triangle with sides 3 cm, 4 cm, and 8 cm.

Draw the base of 3 cm

Draw an arc of radius 4 cm from one end

Draw an arc of radius 8 cm from the other end

The arcs do NOT meet. No triangle can be formed.

**Attempt 2**: Try to construct a triangle with sides 2 cm, 3 cm, and 6 cm.

The arcs just touch at one point or do not meet at all

No proper triangle is formed.

**Conclusion**: Triangles do NOT exist for all sets of side lengths. There are restrictions.

The Problem: Understanding Why Some Lengths Don't Work

Imagine traveling between three cities arranged in a triangle:

**Direct path** between two cities

**Roundabout path** via a third city

Basic principle: A direct path is always shorter than any roundabout path.

**Example with impossible triangle (10 cm, 15 cm, 30 cm):**

Let's assume this triangle exists and has vertices A, B, C where:

AB = 15 cm

AC = 30 cm

BC = 10 cm

Now test the paths between different pairs of vertices:

**Between B and C:**

Direct path BC = 10 cm

Roundabout path B→A→C = 15 + 30 = 45 cm

Direct path IS shorter ✓ (10 < 45)

**Between A and B:**

Direct path AB = 15 cm

Roundabout path A→C→B = 30 + 10 = 40 cm

Direct path IS shorter ✓ (15 < 40)

**Between C and A:**

Direct path CA = 30 cm

Roundabout path C→B→A = 10 + 15 = 25 cm

Direct path is LONGER ✗ (30 > 25)

**This is impossible!** A direct path can never be longer than a roundabout path. Therefore, a triangle with sides 10 cm, 15 cm, and 30 cm cannot exist.

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TRIANGLE INEQUALITY

Definition of Triangle Inequality

**Triangle Inequality Statement**: In a triangle, the sum of any two sides must be greater than the third side.

In mathematical form, for a triangle with sides a, b, and c:

a + b > c

b + c > a

a + c > b

All three conditions must be satisfied simultaneously.

How to Check if a Triangle Can Exist

**Method**: For any three given lengths, check each of the three conditions. If even ONE condition fails, the triangle cannot exist.

Worked Examples

**Example 1: Sides 3 cm, 4 cm, 5 cm**

Check all three conditions:

3 + 4 > 5 → 7 > 5 ✓ (True)

4 + 5 > 3 → 9 > 3 ✓ (True)

3 + 5 > 4 → 8 > 4 ✓ (True)

All three conditions are satisfied. **A triangle exists.**

**Example 2: Sides 2 cm, 3 cm, 6 cm**

Check all three conditions:

2 + 3 > 6 → 5 > 6 ✗ (False)

3 + 6 > 2 → 9 > 2 ✓ (True)

2 + 6 > 3 → 8 > 3 ✓ (True)

The first condition fails. **No triangle exists.**

**Example 3: Sides 10 cm, 15 cm, 30 cm**

Check all three conditions:

10 + 15 > 30 → 25 > 30 ✗ (False)

15 + 30 > 10 → 45 > 10 ✓ (True)

10 + 30 > 15 → 40 > 15 ✓ (True)

The first condition fails. **No triangle exists.**

Important Insight About Checking

When checking the triangle inequality, we only need to check if the **longest side is less than the sum of the other two sides**.

Why? Because:

If the longest side is less than the sum of the other two sides, the other two conditions will automatically be satisfied.

The other two sides are shorter, so their sums will definitely be greater than the longest side.

**Quick Check Method**: Order the sides from smallest to largest. If the longest side < sum of the other two sides, a triangle exists.

**Example**: For 4 cm, 5 cm, 8 cm

Longest side = 8 cm

Sum of other two = 4 + 5 = 9 cm

Is 8 < 9? Yes ✓

**Triangle exists**

Real-Life Application (Indian Context)

In Indian carpentry and construction, when joining three pieces of wood to form a triangular frame, the builder must ensure that no one piece is too long. If one piece is longer than the sum of the other two, they cannot be joined to form a closed triangle. This is a practical application of the triangle inequality.

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VISUALIZING CIRCLES AND TRIANGLE EXISTENCE

Understanding Circle Intersections

When we construct a triangle with sides a, b, and c:

1. We draw the longest side as the base

2. We construct circles (or arcs) with the other two sides as radii

3. The circles must intersect at exactly two points (or touch at one point in the limiting case)

Three Cases of Circle Behavior

**Case 1: Circles Touch at Exactly One Point**

When: Sum of two radii = Base length

Relation: a + b = c (where c is the longest side)

Result: No triangle is formed (degenerate case)

Example: Sides 3 cm, 4 cm, 7 cm

Check: 3 + 4 = 7 (circles touch, no triangle)

**Case 2: Circles Do Not Intersect**

When: Sum of two radii < Base length

Relation: a + b < c

Result: No triangle is formed

Example: Sides 2 cm, 3 cm, 6 cm

Check: 2 + 3 = 5 < 6 (circles don't intersect, no triangle)

**Case 3: Circles Intersect at Two Points**

When: Sum of two radii > Base length

Relation: a + b > c

Result: Triangle is formed

Example: Sides 3 cm, 4 cm, 5 cm

Check: 3 + 4 = 7 > 5 (circles intersect at two points, triangle exists)

Connection Between Circle Intersections and Triangle Inequality

**Key Insight**: A triangle exists if and only if the circles (constructed using the two shorter sides as radii and the longest side as the base) intersect internally.

This happens exactly when the sum of the two shorter sides is greater than the longest side—which is the triangle inequality condition!

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SUMMARY: COMPLETE PROCEDURE FOR CHECKING TRIANGLE EXISTENCE

**To check if a triangle exists for three given lengths:**

1. **Identify the longest side**

2. **Write the inequality**: Longest side < Sum of the other two sides

3. **Check the inequality**:

If TRUE → Triangle exists, and we can construct it

If FALSE → Triangle does not exist

4. **Do we need to check all three triangle inequality conditions?**

No, checking only the longest side is sufficient

The other two conditions will automatically be satisfied

Worked Example

**Question**: Can a triangle exist with sides 4 cm, 5 cm, and 8 cm?

**Solution**:

Arrange in order: 4, 5, 8

Longest side = 8 cm

Sum of other two = 4 + 5 = 9 cm

Check: Is 8 < 9? Yes ✓

**Answer**: Triangle exists

We can also construct it to verify:

Draw base AB = 8 cm

Arc from A with radius 4 cm

Arc from B with radius 5 cm

Arcs intersect at point C

Triangle ABC is formed

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7.3 CONSTRUCTION OF TRIANGLES WHEN SOME SIDES AND ANGLES ARE GIVEN

In many situations, we don't have all three side lengths. Instead, we have some sides and some angle measures. This section covers two important cases.

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CASE 1: TWO SIDES AND THE INCLUDED ANGLE (SAS)

What is "Included Angle"?

The **included angle** is the angle formed between two given sides. It is the angle at the common vertex of the two sides.

**Examples of included angles:**

If sides are 4 cm and 3 cm, and the angle between them is 60°, then 60° is the included angle

If sides are 5 cm and 6 cm, and the angle between them is 30°, then 30° is the included angle

Construction of Triangle ABC with AB = 5 cm, AC = 4 cm, and ∠A = 45°

Here, ∠A (the angle at vertex A) is the included angle between sides AB and AC.

**Step-by-Step Construction:**

**Step 1**: Draw the base AB of length 5 cm using a ruler.

Mark point A at one end

Mark point B at a distance of 5 cm from A

This is one of the given sides

**Step 2**: Construct the angle ∠A = 45° at point A.

Place the center of a protractor at point A

Align the baseline of the protractor with line AB

Mark a point at 45° angle

Draw a ray from A through this marked point

This ray represents the other arm of the 45° angle

**Step 3**: On this ray (the second arm of angle A), measure and mark point C such that AC = 4 cm.

Use a ruler to measure 4 cm from point A along the ray

Mark this point as C

**Step 4**: Join BC using a ruler to complete the triangle.

Draw a straight line from B to C

Triangle ABC is now complete with all required measurements

Why This Construction Works

• AB = 5 cm (constructed) ✓

• AC = 4 cm (measured on the ray) ✓

• ∠A = 45° (constructed at point A) ✓

All three given conditions are satisfied.

Does a Triangle Always Exist for Two Sides and an Included Angle?

**Answer**: Yes, a triangle always exists when two sides and the included angle are given.

**Reason**:

The two sides have definite positive lengths

The angle is between 0° and 180°

These conditions always allow a unique triangle to be formed

There are no restrictions like in the triangle inequality

Unlike the three-sides case, we don't need to check any conditions. The triangle is always possible.

Real-Life Application

In Indian surveying and land measurement, surveyors often measure one side of a plot (along the road) and two angles. Using this information and the SAS construction method, they can determine the complete shape of the land plot.

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CASE 2: TWO ANGLES AND THE INCLUDED SIDE (ASA)

What is "Included Side"?

The **included side** is the side that connects the two vertices where the given angles are located.

**Understanding with an example:**

If ∠A = 45° and ∠B = 80° are given, the included side is AB

The side AB is part of both angles (it forms one arm of ∠A and one arm of ∠B)

Construction of Triangle ABC with AB = 5 cm, ∠A = 45°, and ∠B = 80°

**Step-by-Step Construction:**

**Step 1**: Draw the base AB = 5 cm using a ruler.

Mark point A

Mark point B at a distance of 5 cm from A

**Step 2**: Construct angle ∠A = 45° at point A.

Place the protractor center at A

Align the baseline with line AB

Mark 45° and draw a ray from A

This ray extends upward from A

**Step 3**: Construct angle ∠B = 80° at point B.

Place the protractor center at B

Align the baseline with line AB (but from B's direction)

Mark 80° and draw a ray from B

This ray extends upward from B

**Step 4**: The point where the two rays meet is vertex C.

The ray from A and the ray from B intersect at a point

Call this intersection point C

**Step 5**: Mark point C clearly. Triangle ABC is now complete.

Why This Construction Works

• AB = 5 cm (constructed) ✓

• ∠A = 45° (constructed at point A) ✓

• ∠B = 80° (constructed at point B) ✓

All three given conditions are satisfied.

Can We Find the Third Angle?

Yes! In any triangle, the sum of all three angles = 180°

∠C = 180° - ∠A - ∠B = 180° - 45° - 80° = 55°

Does a Triangle Always Exist for Two Angles and an Included Side?

**Answer**: NO. A triangle does NOT always exist.

**Condition for triangle to exist**: The sum of the two given angles must be less than 180°

**Mathematical condition**: ∠A + ∠B < 180°

**Reason**: The third angle ∠C must have a positive measure:

∠C = 180° - ∠A - ∠B

For ∠C to be positive: 180° - ∠A - ∠B > 0°

This means: ∠A + ∠B < 180° ✓

When Does a Triangle NOT Exist?

If ∠A + ∠B ≥ 180°, then the triangle cannot be formed.

**Example 1**: ∠A = 100°, ∠B = 95°, AB = 6 cm

Sum = 100° + 95° = 195° > 180°

∠C would be = 180° - 195° = -15° (negative angle—impossible)

No triangle exists

**Example 2**: ∠A = 100°, ∠B = 80°, AB = 5 cm

Sum = 100° + 80° = 180°

∠C would be = 180° - 180° = 0° (zero angle—degenerate case)

No triangle exists (three points are collinear)

**Example 3**: ∠A = 60°, ∠B = 70°, AB = 5 cm

Sum = 60° + 70° = 130° < 180° ✓

∠C = 180° - 130° = 50° (positive angle)

Triangle exists ✓

Visualizing Non-Existence

When the two given angles are too large:

The rays drawn from A and B at these angles are nearly parallel or pointing away from each other

They never meet, or meet on the wrong side of AB

No proper triangle is formed

Real-Life Application

In Indian navigation and mapmaking, cartographers often work with angle measurements at known points. Using the ASA method, they can determine the location of distant points (like a ship's position given angle measurements from two coastal lighthouse stations).

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SUMMARY TABLE: TRIANGLE CONSTRUCTIONS

| Given Information | Always Possible? | Condition for Existence | Construction Method |

|---|---|---|---|

| Three sides (SSS) | NO | Triangle Inequality: Each side < sum of other two | Draw base, arcs from endpoints |

| Two sides, included angle (SAS) | YES | None (always exists) | Draw base, angle, mark second side on ray |

| Two angles, included side (ASA) | NO | Sum of angles < 180° | Draw base, angles at endpoints |

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COMMON MISTAKES TO AVOID

1. **Assuming all three side lengths form a triangle**: Always check triangle inequality before constructing.

2. **Forgetting to check if angle sum < 180°**: When given two angles and a side, always verify this condition.

3. **Mixing up "included angle" and "included side"**:

Included angle: the angle between two given sides

Included side: the side between two given angles

4. **Not measuring accurately during construction**: Small errors in compass or protractor use can lead to incorrect triangles.

5. **Assuming there's only one answer**: When constructing with compass arcs, remember that two arcs can intersect at two points. Both points give valid triangles (congruent to each other).

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IMPORTANT DEFINITIONS AND TERMS

**Vertex (Vertices)**: The corner points of a triangle. Three vertices form one triangle.

**Side**: A line segment connecting two vertices. A triangle has three sides.

**Angle**: The amount of rotation between two sides at a vertex.

**Equilateral Triangle**: Triangle with all three sides of equal length.

**Isosceles Triangle**: Triangle with exactly two sides of equal length.

**Scalene Triangle**: Triangle with all three sides of different lengths.

**Included Angle**: The angle formed between two given sides at their common vertex.

**Included Side**: The side that is common to two given angles.

**Triangle Inequality**: The property that the sum of any two sides must be greater than the third side.

**Degenerate Triangle**: When three vertices are collinear (on the same straight line), forming no proper triangle.

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KEY FORMULAS AND RULES

**Triangle Inequality (for sides a, b, c)**:

a + b > c

b + c > a

a + c > b

All three must be true simultaneously.

**Sum of Angles in a Triangle**:

∠A + ∠B + ∠C = 180°

**Quick Check for Triangle Existence** (when sides are ordered as a ≤ b ≤ c):

Triangle exists if and only if a + b > c

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PRACTICE PROBLEMS WITH SOLUTIONS

Problem 1: Check if Triangle Exists

**Question**: Can sides 7 cm, 10 cm, and 15 cm form a triangle?

**Solution**:

Order: 7, 10, 15

Longest side = 15 cm

Sum of other two = 7 + 10 = 17 cm

Check: 15 < 17? Yes ✓

**Answer**: Triangle exists

Problem 2: Find Missing Side Range

**Question**: Sides 5 cm and 8 cm are given. What are all possible lengths for the third side?

**Solution**:

Let the third side be x.

Apply triangle inequality:

5 + 8 > x → x < 13

5 + x > 8 → x > 3

8 + x > 5 → x > -3 (always true for positive x)

Therefore: 3 < x < 13

**Answer**: The third side can be any length strictly between 3 cm and 13 cm. For example: 4, 5, 6, 7, 8, 9, 10, 11, or 12 cm.

Problem 3: Triangle Construction Check

**Question**: Is it possible to construct a triangle with angles 45°, 60°, and an included side of 6 cm?

**Solution**:

∠A = 45°, ∠B = 60°

Sum = 45° + 60° = 105° < 180° ✓

Third angle = 180° - 105° = 75° ✓

**Answer**: Yes, triangle exists and can be constructed

**Construction steps**:

1. Draw AB = 6 cm

2. Construct 45° at A

3. Construct 60° at B

4. Mark intersection as C

Problem 4: Identify Triangle Type

**Question**: Construct a triangle with sides 7 cm, 7 cm, and 9 cm. What type is it?

**Solution**:

Check triangle inequality: 7 + 7 = 14 > 9 ✓

**Type**: Isosceles triangle (two sides are equal: 7 = 7)

**Construction**:

1. Draw base AB = 9 cm

2. Arc from A with radius 7 cm

3. Arc from B with radius 7 cm

4. Mark intersection as C

Problem 5: Angle Sum Application

**Question**: In a triangle, two angles are 50° and 75°. What is the third angle?

**Solution**:

∠A = 50°, ∠B = 75°

Sum of given angles = 50° + 75° = 125°

Third angle ∠C = 180° - 125° = 55°

**Answer**: The third angle is 55°

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IMPORTANT THEOREMS AND PROPERTIES

**Theorem 1 - Triangle Inequality Theorem**

In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.

**Theorem 2 - Angle Sum Property**

The sum of all three interior angles of any triangle is always 180°.

**Theorem 3 - Existence with SSS** (Three Sides Given)

A triangle with given side lengths a, b, c exists if and only if:

a + b > c AND

b + c > a AND

a + c > b

**Theorem 4 - Existence with SAS** (Two Sides and Included Angle)

A triangle always exists when two sides and the included angle are given, provided both sides are positive and the angle is between 0° and 180°.

**Theorem 5 - Existence with ASA** (Two Angles and Included Side)

A triangle with two angles ∠A, ∠B and included side AB exists if and only if:

∠A > 0° AND ∠B > 0°

∠A + ∠B < 180°

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GEOMETRY CONSTRUCTION TOOLS AND THEIR USE

**Ruler**: Used to draw straight lines and measure distances.

**Compass**: Used to draw circles and arcs, and to mark equal distances.

**Protractor**: Used to measure and construct angles.

Basic Compass Skills Needed

• **Opening compass to exact width**: Place needle on zero of ruler and pencil at required measurement

• **Drawing arcs**: Keep needle fixed, rotate pencil around it to create curved line

• **Finding intersection**: Two arcs intersect where they cross

Basic Protractor Skills Needed

• **Aligning protractor**: Center dot on vertex, baseline on one ray of angle

• **Reading angle**: Find required degree marking where second ray passes

• **Constructing angle**: Mark the angle degree, then draw ray through that mark

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REAL-LIFE APPLICATIONS IN INDIA

Architecture

Indian temples often use triangular architectural elements. The stability of these structures depends on the triangle inequality. Pillars supporting triangular roof trusses must satisfy this property.

Agriculture

Farmers marking triangular fields: If the farmer measures three boundaries (three sides), they must satisfy triangle inequality to form a closed field. If not, the field boundaries don't form a closed triangle.

Navigation

Ships navigating Indian coastal waters use lighthouse positions. By measuring angles from two lighthouses and knowing their distance, navigators use ASA construction to locate their position.

Surveying

Indian land surveyors often work with two adjacent sides of a plot and the angle between them (SAS method) to determine the complete boundary of a property.

Carpentry

Carpenters in India constructing triangular roof frames must ensure that no one wooden piece is longer than the sum of the other two pieces, otherwise they cannot be joined properly.

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FREQUENTLY ASKED QUESTIONS

**Q1: Do I need to check all three triangle inequality conditions?**

A: No. Checking if the longest side is less than the sum of the other two is sufficient. The other two conditions will automatically be satisfied.

**Q2: What happens if two sides of a triangle are equal?**

A: It becomes an isosceles triangle. The triangle inequality still applies, and it can always be constructed if it satisfies the inequality.

**Q3: Can angles be given as decimal values like 45.5°?**

A: Yes, angles can be any value between 0° and 180°. Decimal angles can be constructed using a protractor with fine markings.

**Q4: In SAS construction, which side should I draw first?**

A: Any of the two given sides can be drawn first. However, it's conventional to draw one side as the base to make the construction clear.

**Q5: What if the two angle rays in ASA construction don't meet?**

A: This happens when the sum of the two angles equals or exceeds 180°. In this case, a triangle does not exist for those measurements.

**Q6: Can I construct a right triangle using these methods?**

A: Yes. A right triangle has one angle of 90°. You can use any of the three methods:

SSS: If you know all three sides

SAS: If you know two sides and the 90° angle

ASA: If you know the 90° angle and another angle

**Q7: What is the difference between a triangle and a degenerate triangle?**

A: A proper triangle has three non-collinear vertices forming a closed shape. A degenerate triangle has three collinear vertices (on the same line), which is not considered a triangle.

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WORKED EXAMPLES FOR EXAM PREPARATION

Example 1: Complete Problem with All Steps

**Question**: Check if a triangle exists with sides 12 cm, 18 cm, and 25 cm. If yes, construct it. If no, explain why.

**Solution**:

**Step 1: Check Triangle Inequality**

Order sides: 12, 18, 25

Longest side = 25 cm

Sum of other two = 12 + 18 = 30 cm

Check: Is 25 < 30? Yes ✓

Conclusion: Triangle exists

**Step 2: Construct the Triangle**

Construction steps:

1. Draw base AB = 25 cm using a ruler

2. Open compass to 12 cm width

3. Place compass point on A and draw an arc above AB

4. Change compass width to 18 cm

5. Place compass point on B and draw an arc intersecting the first arc

6. Mark the intersection point as C

7. Join AC and BC with straight lines

8. Label the vertices A, B, C

**Step 3: Verification**

AB = 25 cm (constructed)

AC = 12 cm (compass radius from A)

BC = 18 cm (compass radius from B)

All three sides are correctly drawn

**Answer**: Triangle exists and is successfully constructed.

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Example 2: Finding Third Side Range

**Question**: Two sides of a triangle are 6 cm and 10 cm. Find all possible lengths for the third side.

**Solution**:

Let the third side be x cm.

**Apply Triangle Inequality** (all three conditions):

**