Constructions and Tilings

CHAPTER 6: CONSTRUCTIONS AND TILINGS

COMPREHENSIVE CHAPTER NOTES

Class 7 Ganita Prakash (NCF 2023)

---

6.1 GEOMETRIC CONSTRUCTIONS

INTRODUCTION TO GEOMETRIC CONSTRUCTIONS

**Definition**: A geometric construction is the process of drawing geometric figures accurately using only a ruler (unmarked) and a compass, without relying on measurements or freehand drawing.

**Key Principle**: We use only two tools for constructions:

An **unmarked ruler** (straight edge without measurements)

A **compass** (for drawing arcs and circles)

**Why Constructions Matter**:

Constructions help us create exact, symmetrical figures that follow mathematical principles. Instead of drawing shapes approximately, we use properties like congruence and equidistance to ensure accuracy.

---

THE EYE CONSTRUCTION AND PERPENDICULAR BISECTORS

Understanding the Eye Construction (Grade 6 Review)

An eye can be drawn freehand, but for a **symmetrical eye**, we need:

A lower arc and an upper arc of equal radii

Both arcs centered at points that are equidistant from the endpoints of a supporting line

**Supporting line XY**: This is a horizontal line that serves as the base for constructing the eye, though it is not part of the final figure.

**Centers A and B**:

Point A is the center of the upper arc

Point B is the center of the lower arc

For symmetry: AX = AY = BX = BY

**Condition for Symmetry**: The eye will be perfectly symmetrical if and only if the line joining the arc centers (AB) is perpendicular to the supporting line (XY) and passes through its midpoint.

The Perpendicular Bisector

**Definition**: A **perpendicular bisector** of a line segment is a line that:

1. Divides the line segment into two equal parts (bisection)

2. Is perpendicular (at 90°) to the line segment

**Bisection**: The division of any line or geometric object into two identical parts.

Why AB is the Perpendicular Bisector of XY: Congruence Proof

**Given Information**:

AX = AY (A is equidistant from X and Y)

BX = BY (B is equidistant from X and Y)

O is the intersection point of lines AB and XY

**To Prove**: O is the midpoint of XY and AB ⊥ XY

**Step-by-Step Reasoning**:

**Step 1**: Show that ∆ABX ≅ ∆ABY

AX = AY (given)

BX = BY (given)

AB = AB (common side)

By SSS congruence: ∆ABX ≅ ∆ABY

**Step 2**: Use corresponding parts of congruent triangles

From the congruence above: ∠XAB = ∠YAB

Therefore: ∠XAO = ∠YAO (O lies on AB)

**Step 3**: Show that ∆AOX ≅ ∆AOY

AX = AY (given)

∠XAO = ∠YAO (from Step 2)

AO = AO (common side)

By SAS congruence: ∆AOX ≅ ∆AOY

**Step 4**: Draw conclusions from congruence

From the congruence: OX = OY (O is the midpoint of XY)

From the congruence: ∠AOX = ∠AOY

**Step 5**: Prove perpendicularity

∠AOX + ∠AOY = 180° (they form a straight angle)

Since ∠AOX = ∠AOY, we have: 2∠AOX = 180°

Therefore: ∠AOX = ∠AOY = 90°

Thus: AB ⊥ XY

**Conclusion**: AB is indeed the perpendicular bisector of XY.

Creating Different Eye Shapes

**Key Observation**: Any point that has the same distance from X and Y must lie on the perpendicular bisector of XY.

**Why?** If CX = CY and DX = DY, then C and D satisfy the equidistance property. Since XY has only one perpendicular bisector (the line AB), both C and D must lie on this line.

**Application**: We can create eyes of different shapes by:

1. Choosing different pairs of points C and D on the perpendicular bisector AB

2. Using different points as centers for the upper and lower arcs

3. Varying the radii of the arcs

This gives us various eye shapes while maintaining perfect symmetry.

---

CONSTRUCTION OF PERPENDICULAR BISECTOR

Step-by-Step Method Using Ruler and Compass

**Given**: A line segment XY

**Required**: To construct the perpendicular bisector of XY using only an unmarked ruler and compass

**Procedure**:

**Step 1**: Draw intersecting arcs ABOVE the line

Open the compass to any convenient radius (more than half the length of XY)

Center the compass at point X and draw an arc above the line XY

Without changing the radius, center the compass at point Y and draw another arc above XY

These two arcs must intersect; mark this intersection point as A

**Step 2**: Draw intersecting arcs BELOW the line

Using the same radius, center the compass at point X and draw an arc below XY

Without changing the radius, center the compass at point Y and draw another arc below XY

Mark the intersection point of these arcs as B

**Step 3**: Draw the perpendicular bisector

Using an unmarked ruler, join points A and B with a straight line

The line AB is the perpendicular bisector of XY

**Why This Works**:

All points on the circle centered at X are equidistant from X

All points on the circle centered at Y are equidistant from Y

The intersection points A and B satisfy: AX = AY = BX = BY

Any line joining two such points gives the perpendicular bisector

Important Observations About Perpendicular Bisector Construction

**Question 1**: Is it necessary to use the same radius for arcs above and below XY?

**Answer**: **No, it is not necessary.**

**Justification**: We only need to find two points that are equidistant from X and Y. These points can be found using different radii, as long as we use the same radius for the pair of arcs from X and Y at each step.

**Exploration**:

You can use one radius for the upper pair of arcs and a different radius for the lower pair

Both pairs will still give you points equidistant from X and Y

Any two such points determine the perpendicular bisector

Joining any two points on the perpendicular bisector still gives the complete line

**Question 2**: Must we construct both pairs of arcs (above and below XY)?

**Answer**: **No, we only need one pair.**

**Justification**:

Any two points equidistant from X and Y determine the perpendicular bisector

If we have points A and B such that AX = AY and BX = BY, the line AB is the perpendicular bisector

We can construct both pairs of arcs on the same side of XY and still get two valid points

**Question 3**: When drawing intersecting arcs from A and B, must we use the same radii?

**Answer**: **Yes, the radii must be equal for finding points equidistant from X and Y.**

**Justification**:

If we use different radii from X and Y, the intersection points won't be equidistant from both X and Y

Points equidistant from X and Y must lie on the perpendicular bisector

Therefore, equal radii are essential for this construction

---

CONSTRUCTION OF 90° ANGLE AT A GIVEN POINT

Method for Constructing Right Angles

**Given**: A line with a point O marked on it

**Required**: To construct a 90° angle at O using ruler and compass

**Key Idea**: If we can make O the midpoint of a segment XY, then the perpendicular bisector of XY through O will give us a 90° angle.

**Procedure**:

**Step 1**: Mark equidistant points on the line

Using a compass, set any convenient radius

Center at O and mark point X on one side of O at this distance

Without changing the radius, mark point Y on the other side of O at the same distance

Now O is the midpoint of XY

**Step 2**: Construct the perpendicular bisector

Using the same method as before, construct arcs above and below the segment XY

From X and Y, draw intersecting arcs at a point A (above the line)

Draw another pair of intersecting arcs at point B (below the line)

Join A and B with a ruler

**Step 3**: The perpendicular bisector is the 90° angle

The line AB passes through O and is perpendicular to the original line

This creates a 90° angle at O

**Optimization**: Since we already know O lies on the perpendicular bisector, we only need one pair of intersecting arcs (either above or below) to determine the direction of the perpendicular bisector. The second point on this line is already known (O itself).

**Result**: The angle formed between the original line and the constructed perpendicular is exactly 90°.

---

ANCIENT INDIAN CONSTRUCTIONS: ŚULBA-SŪTRAS

Historical Context

**Śulba-Sūtras**: These are ancient geometric texts from the Vedic period of India, meaning "rope instructions" or "rules of the rope."

**Purpose**: The Śulba-Sūtras contain mathematical and geometric procedures for constructing fire altars (agni) used in Vedic rituals.

**Part of Vedāṅgas**: The Śulba-Sūtras are one of the six Vedāṅgas, which literally means "limbs of the Vedas" - supplementary texts explaining aspects of Vedic knowledge.

**Key Authors**: Important Śulba-Sūtras were composed by:

Āpastamba

Kātyāyana

And others

Tools Used in Śulba-Sūtras Constructions

**Rope as a Geometric Tool**: Instead of modern compass and ruler, ancient Indian mathematicians used ropes, which could:

Be stretched to form straight lines

Be used to draw circles and arcs

Be folded to find midpoints

Be fastened at specific points

**Advantages of Rope**:

Could create very large constructions (for ground-level altar construction)

Practical for outdoor use

Mathematically equivalent to modern compass and ruler constructions

Rope Construction of Perpendicular Bisector (Kātyāyana-Śulbasūtra 1.2)

**Given**: A line segment XY drawn on the ground

**Required**: To construct the perpendicular bisector using a rope

**Procedure**:

**Step 1**: Set up the supports

Fix a small pole or peg vertically at point X

Fix another pole or peg vertically at point Y

These will serve as anchor points

**Step 2**: Prepare the rope

Take a sufficiently long rope

Make two loops at the ends of the rope

Fold the rope into half (without counting the looped portions)

Mark the midpoint of the rope carefully

**Step 3**: First construction (above the line)

Fasten the looped ends of the rope to the poles at X and Y

Pull the midpoint of the rope upward (above XY)

Ensure both parts of the rope on either side are fully stretched and tight

Mark this position as point A

**Step 4**: Second construction (below the line)

Keep the rope fastened to the same poles

Pull the midpoint of the rope downward (below XY)

Again ensure both rope segments are fully stretched

Mark this position as point B

**Step 5**: Draw the perpendicular bisector

The line connecting points A and B is the perpendicular bisector

**Why This Works**:

When the rope is folded in half, both halves have equal length

When stretched from X to A and from Y to A, both rope segments have the same length (being halves of the whole rope)

Therefore: AX = AY

Similarly: BX = BY

Points A and B both satisfy the equidistance property, so AB is the perpendicular bisector

**Mathematical Equivalence**: This rope method is mathematically identical to the compass-and-ruler method, using the same principle of equidistance.

Rope Construction of 90° Angle

**Question**: Can we construct a 90° angle at a point using only a rope?

**Answer**: Yes, using similar methods:

1. Mark three points on the rope at distances forming a 3-4-5 right triangle

2. Fold the rope to form a triangle with these proportions

3. The angle formed will be 90°

**Historical Note**: The 3-4-5 relationship (Pythagorean relationship) was known to ancient Indian mathematicians and appears in the Śulba-Sūtras.

---

ANGLE BISECTION

Introduction to Angle Bisectors

**Definition**: The **angle bisector** is a ray or line that divides an angle into two equal parts.

**Purpose in Design**: When we need to create symmetric or repeated patterns with equal angles, angle bisection is essential.

The 8-Petal Figure Example

**Context**: Consider a flower-like design with 8 equal petals arranged in a circle.

**Constraint**: For the petals to be equally spaced, the angle between any two adjacent petals must be equal.

**Calculation of Angle**:

Complete angle around a point = 360°

Number of petals = 8

Angle between adjacent petals = 360° ÷ 8 = 45°

**Construction Challenge**: How do we construct a 45° angle using only ruler and compass?

**Solution**:

1. First construct a 90° angle (using perpendicular bisector method)

2. Then bisect the 90° angle to get 45°

Method to Bisect Any Angle

**Given**: An angle ∠XOY

**Required**: To construct the angle bisector of ∠XOY

**Mathematical Foundation Using Congruence**:

If we can construct triangles ∆OAC and ∆OBC such that:

∆OAC ≅ ∆OBC

Then the corresponding angles will be equal:

∠AOC = ∠BOC

This means OC bisects ∠AOB.

**What makes these triangles congruent?**

By SSS (Side-Side-Side) congruence, if:

OA = OB (equal distances from O)

AC = BC (equal distances between A and B from point C)

OC = OC (common side)

Then ∆OAC ≅ ∆OBC, making OC the angle bisector.

Step-by-Step Angle Bisection Procedure

**Step 1**: Mark points at equal distance from the vertex

```

Given angle ∠XOY at vertex O

Mark points A and B such that OA = OB

(These points lie on the arms of the angle, at equal distances from O)

```

**Method for Step 1**:

Open compass to any convenient radius

Center at O and mark an arc on ray OX, marking point A

Keep the same radius and mark an arc on ray OY, marking point B

Now OA = OB

**Step 2**: Find the equidistant point from A and B

```

Using any sufficiently long radius (more than half of AB):

Center compass at A and draw an arc

Without changing radius, center at B and draw another arc

These arcs intersect at point C

Now we have AC = BC

```

**Step 3**: Draw the angle bisector

```

Using a ruler, join O to C

Line OC is the angle bisector of ∠XOY

```

**Why This Works**:

∆OAC has sides: OA, AC, OC

∆OBC has sides: OB, BC, OC

We know: OA = OB (from Step 1), AC = BC (from Step 2), OC = OC (common)

By SSS congruence: ∆OAC ≅ ∆OBC

Therefore: ∠AOC = ∠BOC (corresponding angles in congruent triangles)

So OC bisects ∠AOB

Important Observations About Angle Bisection

**Question**: If arcs of equal radius are drawn on the other side of line AB instead of on the same side, will the line still be an angle bisector?

**Answer**: **Yes, the line OC will still be the angle bisector.**

**Justification**:

The principle of congruence remains the same

Whether we mark point C above or below AB, we still get AC = BC

The congruence ∆OAC ≅ ∆OBC still holds

Therefore ∠AOC = ∠BOC still holds

The only difference is the direction or orientation of point C

The bisector line remains the same

**Practical Implication**: We have flexibility in where we construct the intersecting arcs, which can be useful depending on the space available or the design requirements.

Angles Constructible Using Angle Bisection

**Starting with 90°**: We can construct 45° by bisecting once.

**Further bisections**:

Bisect 45° to get 22.5°

Bisect 22.5° to get 11.25°

And so on...

**Constructible angles from 90° using repeated bisection**:

45°

22.5°

11.25°

5.625°

And all other angles of the form: 90°/(2^n) where n = 1, 2, 3, ...

**Can we construct 65.5°?**

**Analysis**:

65.5° is not of the form 90°/(2^n)

Using only angle bisection of a single starting angle (90°), we cannot construct 65.5°

However, if we combine multiple angles and their bisections (for example: 45° + 22.5° + some combination), we might be able to construct specific angles

But 65.5° specifically cannot be constructed using only angle bisection of 90°

**Note**: Some angles can be constructed by combining multiple bisections and additions, but this requires more advanced techniques beyond simple bisection.

Constructing a Rope Angle Bisector

**Question**: How can we construct an angle bisector using only a rope?

**Answer**: Using the equidistance principle

**Method**:

1. At the vertex O, mark two points A and B on the arms of the angle using equal rope lengths

2. Cut a rope to connect A and B

3. Fold this rope in half to find its midpoint C

4. The line from O through C bisects the angle

**Justification**:

When rope AB is folded in half at C, we have AC = BC

Combined with OA = OB (from the first step)

This creates the congruent triangle condition

Therefore OC bisects the angle

---

REPEATING ANGLES AND COPYING ANGLES

Creating Designs with Repeating Units

**Design Pattern Example**: A geometric design where a single unit repeats in different orientations.

**Challenge**: Each repeated unit must have:

Exactly the same arm lengths

Exactly the same angles between the arms

**Solution**:

We can ensure equal arm lengths using a compass (measuring arcs)

We need a method to ensure equal angles - this requires **copying angles**

Copying an Angle: Method and Justification

**Given**: An angle at point A with arms AX and AB

**Required**: To create an exact copy of this angle at a different location (point X)

Step-by-Step Angle Copying Procedure

**Step 1**: Create an isosceles triangle from the original angle

```

Draw an arc from vertex A with any convenient radius

This arc intersects arm AX at point B

This arc intersects arm AY at point C

Triangle ABC is isosceles with AB = AC

∠BAC is the angle we want to copy

```

**Step 2**: Reproduce the triangle at the new location

```

At the new location (point X where we want the angle):

Draw an arc of the same radius as in Step 1

Mark a point Z on this arc

This will be equivalent to point B in the original triangle

The arc from X is equivalent to the arc from A

```

**Step 3**: Transfer the base measurement

```

Measure the length BC from the original triangle using compass

Without changing the compass width:

Center at Z and draw an arc intersecting the arc from X

Mark the intersection point as Y

This creates YZ = BC

```

**Step 4**: Connect to form the copied angle

```

Draw a line from X through Y

This line creates ∠ZXY at the new location

By SSS congruence (since triangle XYZ ≅ triangle ABC), we have:

∠ZXY = ∠BAC

```

Why Angle Copying Works: Congruence Justification

**Original Triangle**: ∆ABC where:

∠BAC is the angle to be copied

AB = AC (radius of arc from A)

BC is the distance between the two intersection points

**Copied Triangle**: ∆XYZ where:

XZ is constructed with the same radius as AB

YZ is constructed with the same length as BC

XY will be the direction of the copied angle

**Congruence by SSS**:

AB = XZ (same radius, by construction)

AC = XZ (wait, we need to reconsider)

**Actually, let me correct this**:

**Original Triangle**: ∆ABC where:

A is the vertex of the original angle

B is on one arm (at distance r from A)

C is on the other arm (at distance r from A)

AB = AC = r (same radius from A)

BC is the chord connecting B and C

**Copied Triangle**: ∆XYZ where:

X is the vertex of the new angle

Z is at distance r from X (same radius)

Y is at distance equal to BC from Z

XY is the second arm of the new angle

**Congruence Proof**:

AB = XZ = r (same radius used in both constructions)

AC is not directly used, but we use the chord BC

BC = YZ (by construction, we transfer the compass length BC to ZY)

XZ = XZ (we need to complete this)

**Corrected Triangle Pairs**:

Triangle ABC: A is vertex, AB = AC = r, BC is the chord

Triangle XYZ: X is vertex, XZ = r, and we need YZ = BC

Wait, the original angle is ∠BAC, formed by arms AB and AC.

**Better Approach**:

Original angle ∠BAC at vertex A

An arc from A with radius r intersects the arms at B and C

The triangle formed is isosceles ∆ABC with AB = AC = r and angle ∠BAC

**At new location**:

We create an arc from X with same radius r

This gives us point Z on one arm

We measure BC and mark point Y such that ZY = BC and Y is on the arc from X

Now we have triangle XYZ where:

XZ = AB = r

YZ = BC (by construction)

The angle ∠ZXY = ∠BAC

**Congruence of triangles ABC and XYZ**:

AB = XZ (both equal to r)

BC = YZ (transferred by compass)

AC = XY (both are the connecting sides)

By SSS congruence: ∆ABC ≅ ∆XYZ

Therefore: ∠BAC = ∠ZXY (corresponding angles in congruent triangles)

**Conclusion**: The copied angle equals the original angle.

Application: Constructing Repeating Patterns

Once we can copy angles, we can:

1. Create multiple units with identical angles

2. Arrange them in different orientations

3. Build complex designs with perfect geometric symmetry

---

CONSTRUCTION OF PARALLEL LINES

Relationship Between Parallel Lines and Equal Angles

**Recall from earlier geometry**: When a transversal intersects two parallel lines:

Corresponding angles are equal

If corresponding angles are equal, the lines are parallel

**Construction Principle**: We will use the converse:

If we can make equal corresponding angles, we can construct parallel lines

Method to Construct Parallel Lines Using Ruler and Compass

**Given**:

A line m to which we need a parallel line

A point B through which the parallel line must pass

**Required**: To construct a line through B parallel to m

**Key Construction Elements**:

**Step 1**: Choose a transversal

Draw a line l that passes through point B and intersects line m at point A

Line l will act as a transversal

**Step 2**: Identify the angle to be copied

The angle ∠MAL (or ∠BAL, where the angle is between line m and transversal l)

This angle is at point A

This is the angle we will copy to point B

**Step 3**: Construct equal corresponding angles

At point B on line l, construct an angle equal to ∠MAL

This is done using the angle copying method described earlier:

Draw an arc of equal radius from A and B

Transfer the chord length from A to B

Create the copied angle at B

**Step 4**: Complete the parallel line

Draw a line through B along the copied angle

This line n is parallel to line m

We have m || n

Why This Creates Parallel Lines

**Reason**: If a transversal intersects two lines such that the corresponding angles are equal, then the two lines must be parallel.

**In our construction**:

Line l is the transversal

Angle at A (on line m) = Angle at B (on line n) [by construction]

These are corresponding angles

Therefore, m || n

Practical Considerations

**Choosing the transversal angle**:

We choose the angle between line m and l

We make the same angle between the parallel line n and l

The transversal angle should be chosen for convenience (not too acute, not close to 0° or 90°)

**Accuracy**:

The parallel line will be more accurate if the copied angle is precise

Use a compass with sufficient radius to make the arc measurements clear

---

ARCH DESIGNS

Introduction to Arches

**What is an arch?**: A curved structural element that spans a space and provides support.

**Architectural Importance**: Arches are fundamental in architecture, used in:

Bridges (spanning rivers)

Doorways and windows (supporting weight)

Decorative elements (aesthetic appeal)

**Examples from Indian Architecture**:

**Central Park, New York**: Features beautiful stone arches

**Diwan-i-Aam, Red Fort, Delhi**: Shows intricate arch designs with trefoil and pointed arch styles

Construction Principles for Arches

**Supporting Lines**: Every arch construction requires:

**Base lines**: Line segments that define the width and position of the arch

**Center lines**: Arcs are drawn from carefully positioned centers

**Support lines**: Additional construction lines that ensure symmetry

**Symmetry Requirements**: For aesthetically pleasing arches:

If the arch spans between points A and B, both sides should be mirror images

The arch should be symmetric about a vertical line through its midpoint

This requires equal angles and equal distances

---

TREFOIL ARCH CONSTRUCTION

Understanding the Trefoil Arch

**What is it?**: A trefoil arch consists of three leaf-shaped or petal-shaped lobes meeting at a point.

**Symmetry**: The arch has perfect symmetry:

One central vertical line of symmetry

Both sides are identical mirror images

Construction of Trefoil Arch

**Step 1**: Prepare the base and support lines

```

Mark points A and D on a horizontal line

These represent the width of the arch

```

**Step 2**: Identify symmetric points

```

Mark points B and C such that:

B lies on AD

C lies on AD

The arch is symmetric about the midpoint of AD

For symmetry: AB = CD

```

**Step 3**: Construct equal angles at the ends

```

Construct angles at A and D such that:

∠BAE = ∠CDF (where E and F are directions for the arc centers)

These angles must be equal for symmetry

Use angle construction method to ensure equality

```

**Step 4**: Draw the arcs

```

From the point determined by the angle at A, draw an arc

From the point determined by the angle at D, draw another arc

These arcs form the trefoil shape

Adjust the radii of the arcs as needed for aesthetic appearance

```

Characteristics of Trefoil Arch

**Three lobes**:

One or more curved sections form the characteristic three-part pattern

Can have the widest part at the top (most common) or at different positions

**Application**:

Decorative arches in historical buildings

Window designs in Gothic and Islamic architecture

Structural elements in bridges and fortifications

---

POINTED ARCH CONSTRUCTION

Understanding Pointed Arches

**Description**: A pointed arch comes to a sharp point at the apex (top), unlike rounded arches.

**Structural Advantage**: The pointed shape redirects weight more efficiently downward.

**Architectural Significance**:

Common in Gothic architecture

Found in Islamic and Islamic-influenced architecture (Red Fort, Delhi)

More sophisticated than simple rounded arches

Construction of Pointed Arch

**Step 1**: Prepare support lines

```

Draw two line segments of equal length

These segments serve as the sides of the pointed arch

Position them at an angle (not parallel)

The angle determines the "pointedness" of the arch

```

**Step 2**: Identify the midpoints

```

Find and mark the midpoint of each line segment

These midpoints are crucial for the arch construction

```

**Step 3**: Construct the arcs

```

From the endpoint of one segment, draw an arc

This arc should pass through or near the midpoint of the other segment

Similarly, from the endpoint of the second segment, draw an arc

This arc should pass through or near the midpoint of the first segment

These two arcs meet at a point (the apex)

Creating the characteristic pointed shape

```

**Step 4**: Adjust radii for aesthetics

```

The radius of each arc can be adjusted to:

Make the point sharper (smaller radius) or more rounded

Create different artistic effects

Maintain proper proportions with the supporting structure

```

Connection to "Wavy Wave" (Grade 6 Reference)

The pointed arch construction uses similar principles to the wavy wave pattern from Grade 6:

Both rely on arcs of specific radii

Both use midpoints of segments as reference points

Both create elegant curves from simple support structures

---

REGULAR POLYGONS AND HEXAGONS

Definition of Regular Polygons

**Regular Polygon**: A polygon that has:

1. All sides of equal length

2. All interior angles of equal measure

**Examples**:

**Equilateral triangle** (3 sides): All sides equal, all angles = 60°

**Square** (4 sides): All sides equal, all angles = 90°

**Regular pentagon** (5 sides): All sides equal, all angles = 108°

**Regular hexagon** (6 sides): All sides equal, all angles = 120°

Construction Feasibility

**Can construct easily**:

Equilateral triangles ✓

Squares ✓

Regular hexagons ✓

**More difficult**:

Regular pentagons (requires deeper understanding of triangles and pentagons - postponed to later years)

---

REGULAR HEXAGON CONSTRUCTION

The Key Insight: Hexagons and Equilateral Triangles

**Discovery**: A regular hexagon can be divided into six congruent equilateral triangles, all meeting at a central point.

**Visual Pattern**:

```

C

/ \

/ \

B D

/ \ / \

/ \ / \

A O E F

\ / \ /

\ / \ /

L G

\ /

\ /

H

```

Understanding the Hexagon-Triangle Relationship

**Question**: Do all triangles in this arrangement form equilateral triangles?

**Answer**: Yes, when a regular hexagon is divided by drawing lines from center to all vertices, six congruent equilateral triangles are formed.

**Why?**:

Each central angle = 360° ÷ 6 = 60°

In a regular hexagon, the distance from center to each vertex is equal

If we have two equal radii meeting at 60°, the triangle formed is equilateral (isosceles with 60° vertex angle means all angles are 60°)

Key Angle Theorem

**Complete Angle Around a Point**: The sum of all angles around a single point equals 360°.

**Implication**: If we have multiple angles meeting at a central point, they can completely surround the point without gaps or overlaps if their sum is exactly 360°.

**Example with Mixed Angles**:

```

Suppose we have angles of: 40°, 60°, 50°, 30°,