Symmetry

Given: Figures at the start of the chapter (rangoli/kolam-type patterns) and a picture of a cloud.

Concept: A line of symmetry divides a figure into two mirror-image halves that exactly overlap when folded.

Answer:
- The decorative figures (like rangoli patterns) at the start of the chapter generally have lines of symmetry — they can have multiple lines of symmetry (vertical, horizontal, and diagonal) depending on the specific figure.
- A cloud does not have a line of symmetry because its boundary is irregular and uneven; no fold line will make both halves overlap exactly.

Conclusion: The rangoli/decorative figures have lines of symmetry; the cloud does not.

Given: Several geometric and natural shapes.

Concept: A line of symmetry is a line along which a figure can be folded so that both halves match exactly.

Answer (based on standard shapes typically shown in this exercise):

- Equilateral triangle: 3 lines of symmetry (one from each vertex to the midpoint of the opposite side).
- Square: 4 lines of symmetry (2 along diagonals, 2 along midpoints of opposite sides).
- Rectangle (non-square): 2 lines of symmetry (along midpoints of opposite sides; diagonals are NOT lines of symmetry).
- Isosceles triangle: 1 line of symmetry (the perpendicular bisector of the base).
- Scalene triangle: 0 lines of symmetry.
- Circle: Infinite lines of symmetry (any diameter).
- Regular hexagon: 6 lines of symmetry.
- Irregular figures: 0 lines of symmetry.

Note: Since the actual figures in the image cannot be seen, students should apply the above concept to each shape shown and draw the fold line(s) accordingly.

Given: Square sheets of paper were folded and a hole was punched; after unfolding, the positions of holes are shown in figures (a), (b), (c), (d).

Concept: When a paper is folded along a line of symmetry and a hole is punched, the hole appears symmetrically on both halves when unfolded. The fold line is the line of symmetry between the two holes.

Answer:

- Figure (a): Two holes appear symmetrically about a vertical line through the centre of the square. So the paper was folded along the vertical centre line.

- Figure (b): Two holes appear symmetrically about a horizontal line through the centre. So the paper was folded along the horizontal centre line.

- Figure (c): Two holes appear symmetrically about a diagonal line. So the paper was folded along one of the diagonal lines of the square.

- Figure (d): Only one hole appears, and it lies exactly on the fold line (centre). This means the paper was folded such that the hole was punched exactly on the fold line itself, so when unfolded, only one hole is visible. The paper was folded along the line passing through that hole (e.g., the vertical or horizontal centre line), and the hole was punched right on the fold.

Conclusion: The fold line is always the line of symmetry between the two holes (or through the single hole in case d).

Given: In each figure, one or more holes are shown along with the line(s) of symmetry.

Concept: The mirror image of each hole about the given line of symmetry gives the position of the other hole(s). If a point is at distance $d$ from the line of symmetry, its mirror image is at the same distance $d$ on the other side, along the perpendicular to the line.

Method: For each hole, reflect it across the given line of symmetry to find the corresponding hole.

- Figure (a): One line of symmetry (vertical). Reflect the given hole across the vertical line — the other hole is at the mirror position on the right side.

- Figure (b): One line of symmetry (horizontal). Reflect the given hole across the horizontal line — the other hole is directly below (or above) at the same horizontal distance from the line.

- Figure (c): One line of symmetry (diagonal). Reflect the given hole across the diagonal — the other hole is at the mirror position across the diagonal.

- Figure (d): Two lines of symmetry (vertical and horizontal). Reflect the given hole across both lines — this gives 3 additional holes, one in each of the other three quadrants.

- Figure (e): Two lines of symmetry (both diagonals or vertical+horizontal). Reflect the given hole across both lines of symmetry to find all corresponding holes.

Note: Since the exact positions in the images cannot be seen, students should apply the reflection principle: measure the perpendicular distance of the hole from the line of symmetry and mark the mirror image at the same distance on the other side.

Given: A square sheet is folded (vertically or horizontally) and a cut is made along a straight line.

Concept: When a folded paper is cut and unfolded, the cut shape is reflected about the fold line, producing a symmetric hole.

Answer:

- Figure (a) — Vertical fold, triangular cut at the edge:
When unfolded, the two triangular cuts mirror each other about the vertical fold line, producing a rhombus (diamond) shaped hole.

- Figure (b) — Vertical fold, rectangular/straight cut:
When unfolded, the cut is mirrored, producing a rectangular hole.

- Figure (c) — Horizontal fold, triangular cut:
When unfolded, the triangular cut is mirrored about the horizontal fold line, producing a square or rhombus shaped hole (depending on the angle of cut).

- Figure (d) — Fold along diagonal, straight cut:
When unfolded, the cut is mirrored about the diagonal, producing a square hole (if the cut is perpendicular to the diagonal).

Verification: Students should physically fold and cut the paper to verify their predictions.

Key principle: The shape of the hole = the cut shape + its mirror image about the fold line.

Given: We need to produce a square hole in the centre of a sheet using folds and a single straight cut.

Concept: By folding the paper appropriately, a single straight cut can produce a symmetric shape when unfolded.

Answer:

Part (a) — Square hole with sides parallel to the edges of the paper:

Step 1: Fold the square sheet in half vertically (left half over right half).
Step 2: Fold again horizontally (top half over bottom half). Now the paper is folded into four layers.
Step 3: Make a single straight cut at an angle of $45°$ from the corner that corresponds to the centre of the original sheet.
Step 4: When unfolded, the cut produces a square hole in the centre with sides parallel to the paper's edges.

Part (b) — Square hole tilted at 45° (diamond orientation):

Step 1: Fold the square sheet along one diagonal.
Step 2: Fold again along the other diagonal. Now the paper is in four triangular layers.
Step 3: Make a single straight cut parallel to the open edges.
Step 4: When unfolded, the cut produces a square hole tilted at 45° (diamond shape) in the centre.

Note: Check that the resulting 4-sided figure has all sides equal and all angles $90°$ to confirm it is a square.

Given: Various shapes.

Concept: The number of lines of symmetry of a regular polygon with $n$ sides is $n$.

Answer:

Part (a): (Based on the figure shown — appears to be a shape like a plus/cross or similar)
A plus/cross shape with 4 equal arms has 4 lines of symmetry (2 along the arms, 2 along the diagonals between arms).
*(Note: The exact answer depends on the figure shown; students should apply the fold test.)*

Part (b) — Equilateral triangle (3 equal sides, 3 equal angles):
$$\text{Number of lines of symmetry} = 3$$
The three lines of symmetry go from each vertex to the midpoint of the opposite side.

Part (c) — Regular hexagon (6 equal sides, 6 equal angles):
$$\text{Number of lines of symmetry} = 6$$
Three lines connect opposite vertices, and three lines connect midpoints of opposite sides.

Summary Table:
| Shape | Lines of Symmetry |
|---|---|
| Shape (a) | 4 (if cross/plus shape) |
| Equilateral triangle | 3 |
| Regular hexagon | 6 |

Given: Various figures (geometric shapes, letters, natural objects).

Concept: A line of symmetry divides a figure into two identical mirror halves.

Method: For each figure, imagine folding it along different lines (vertical, horizontal, diagonal) and check if both halves overlap exactly.

General answers for common figures in this type of exercise:

- Arrow pointing right: 1 line of symmetry (horizontal, along the direction of the arrow).
- Letter A: 1 line of symmetry (vertical).
- Letter H: 2 lines of symmetry (vertical and horizontal).
- Regular pentagon: 5 lines of symmetry.
- Parallelogram (non-rectangle): 0 lines of symmetry.
- Kite: 1 line of symmetry (along the main diagonal).
- Semi-circle: 1 line of symmetry (vertical, through the midpoint of the diameter).

Note: Students should trace each figure from the textbook and draw the fold lines directly on the traced figure.

Given: A kolam (traditional South Indian floor pattern).

Concept: A kolam is typically drawn with high symmetry. Lines of symmetry are lines along which the pattern can be folded to produce identical halves.

Answer:
A typical kolam of this type has 4 lines of symmetry:
1. A vertical line through the centre.
2. A horizontal line through the centre.
3. A diagonal line from top-left to bottom-right.
4. A diagonal line from top-right to bottom-left.

Note: The exact number depends on the specific kolam shown. Students should trace the kolam and test each potential fold line.

Given: Requirements for different types of triangles based on lines of symmetry.

Concept: Lines of symmetry in triangles depend on the type of triangle.

Answer:

Part (a) — Triangle with exactly one line of symmetry:
Draw an isosceles triangle (two equal sides). The line of symmetry is the perpendicular bisector of the base (the line from the apex to the midpoint of the base).
$$\text{Example: A triangle with sides } 5\text{ cm}, 5\text{ cm}, 3\text{ cm}$$

Part (b) — Triangle with exactly three lines of symmetry:
Draw an equilateral triangle (all three sides equal, all angles $60°$). Each of the three lines from a vertex to the midpoint of the opposite side is a line of symmetry.
$$\text{Example: A triangle with all sides } 5\text{ cm}$$

Part (c) — Triangle with no line of symmetry:
Draw a scalene triangle (all three sides of different lengths). No fold line will make the two halves overlap.
$$\text{Example: A triangle with sides } 3\text{ cm}, 4\text{ cm}, 5\text{ cm}$$

Is it possible to have exactly two lines of symmetry?
No, it is not possible. A triangle cannot have exactly two lines of symmetry. If a triangle has two lines of symmetry, the third line must also be a line of symmetry, making it equilateral with three lines of symmetry. So a triangle can have only 0, 1, or 3 lines of symmetry.

Given: Figures with curved boundaries and specific numbers of lines of symmetry.

Concept: Curved figures can also have lines of symmetry.

Answer:

Part (a) — Exactly one line of symmetry (with curved boundary):
Draw a semi-circle (half circle with a straight diameter). The vertical line through the midpoint of the diameter is the only line of symmetry.
*Alternatively:* Draw a heart shape — it has exactly 1 vertical line of symmetry.

Part (b) — Exactly two lines of symmetry (with curved boundary):
Draw an ellipse (oval shape). It has exactly 2 lines of symmetry: one along the major axis and one along the minor axis.

Part (c) — Exactly four lines of symmetry (with curved boundary):
Draw a circle inscribed in a square or a 4-petal flower shape. Such a figure has 4 lines of symmetry: 2 along the axes and 2 along the diagonals.
*Alternatively:* A circle has infinite lines of symmetry, so draw a shape like a square with semicircular bumps on each side — this gives exactly 4 lines of symmetry.

Given: Partial figures on squared paper with a blue line of symmetry.

Concept: To complete a figure so that a given line is a line of symmetry, reflect every point/segment of the given part across the line of symmetry.

Method: For each point in the given figure, find its mirror image across the blue line:
- If the blue line is vertical: a point at column $c$ from the line maps to column $c$ on the other side (same row).
- If the blue line is horizontal: a point at row $r$ from the line maps to row $r$ on the other side (same column).
- If the blue line is diagonal: swap the row and column distances from the line.

Answer:

Figure (b): The blue line is vertical. Reflect each coloured square to its mirror position on the right side of the blue line.

Figure (c): The blue line is diagonal (hint: rotate the book). Reflect each square across the diagonal line. *(Tip: rotating the book 45° makes it easier to see the reflection.)*

Figure (d): The blue line is horizontal. Reflect each coloured square to its mirror position below the blue line.

Figure (e): The blue line is vertical. Reflect each coloured square to its mirror position on the other side.

Figure (f): The blue line is diagonal. Reflect each square across the diagonal. *(Tip: rotating the book helps.)*

Key Rule: Every point in the completed figure must be at the same perpendicular distance from the line of symmetry as its mirror image, on the opposite side.

Given: Partial figures on squared paper with two blue lines of symmetry.

Concept: When a figure has two lines of symmetry, every point must have mirror images across both lines. This means reflecting across one line and then the other (which is equivalent to a 180° rotation about the intersection point).

Method:
Step 1: Reflect the given part across the first blue line of symmetry.
Step 2: Reflect the result (and the original) across the second blue line of symmetry.
Step 3: All four parts together form the complete symmetric figure.

Answer:

Figure (a): Two lines of symmetry (vertical and horizontal). Reflect the given portion into all four quadrants.

Figure (b): Two lines of symmetry. Apply reflections across both lines to complete the figure.

Figure (c): Two lines of symmetry. Complete by reflecting across both axes.

Figure (d): Two lines of symmetry. Reflect the given squares into the remaining three sections.

Figure (e): Two lines of symmetry. Complete the figure by reflecting across both lines.

Figure (f): Two lines of symmetry. Apply both reflections to complete the pattern.

Important: After completing, verify by checking that folding along either blue line makes both halves overlap exactly.

Given: Partial figures on a dot grid.

Concept: Adding two line segments to a figure to create a line of symmetry.

Method:
Step 1: Examine the existing lines in the figure.
Step 2: Identify a potential line of symmetry (vertical, horizontal, or diagonal).
Step 3: Draw two additional line segments such that the completed figure is symmetric about that line.

Answer:
For each figure on the dot grid:
- Identify which line (vertical/horizontal/diagonal) could serve as the line of symmetry.
- Add two line segments that are mirror images of each other about that line, OR add two segments that complete the figure into a recognisable symmetric shape (like a triangle, rectangle, or kite).

Example approach: If the existing figure has two lines going to the right of a central dot, draw two mirror-image lines going to the left of the same dot, making a symmetric 'V' or 'X' shape.

Note: Multiple correct answers are possible. Students should verify by folding (or imagining folding) along the chosen line of symmetry.

Given: Three figures with a marked centre point.

Concept: An angle of symmetry is an angle through which a figure can be rotated about a fixed point so that it looks exactly the same as before. The angles of symmetry are always multiples of the smallest angle of symmetry, and $360°$ is always an angle of symmetry.

Answer:

Figure (a): (Appears to be a figure with 2-fold symmetry, like a rectangle or S-shape)
Smallest angle of symmetry $= 180°$
Angles of symmetry: $\mathbf{180°, 360°}$

Figure (b): (Appears to be a figure with 3-fold symmetry, like a 3-armed shape)
Smallest angle of symmetry $= \dfrac{360°}{3} = 120°$
Angles of symmetry: $\mathbf{120°, 240°, 360°}$

Figure (c): (Appears to be a figure with 4-fold symmetry, like a 4-armed shape or square)
Smallest angle of symmetry $= \dfrac{360°}{4} = 90°$
Angles of symmetry: $\mathbf{90°, 180°, 270°, 360°}$

Note: Students should verify by tracing the figure and rotating it about the marked point.

Given: Seven figures.

Concept: A figure has more than one angle of symmetry if it can be rotated by at least two different angles (both strictly between $0°$ and $360°$, plus $360°$ itself) and look the same. This happens when the figure has rotational symmetry of order 2 or more.

Answer:
Figures that have rotational symmetry of order $\geq 2$ have more than one angle of symmetry.

- Figures that look the same after rotation by $180°$ (order 2): have angles of symmetry $180°$ and $360°$ — more than one angle of symmetry ✓
- Figures that look the same after rotation by $120°$ (order 3): have angles $120°, 240°, 360°$ — more than one angle of symmetry ✓
- Figures that look the same after rotation by $90°$ (order 4): have angles $90°, 180°, 270°, 360°$ — more than one angle of symmetry ✓
- Figures with no rotational symmetry (order 1): only $360°$ is an angle of symmetry — only one angle of symmetry ✗

Based on typical figures in this exercise:
- A square: more than one angle of symmetry (90°, 180°, 270°, 360°) ✓
- A regular triangle: more than one angle of symmetry (120°, 240°, 360°) ✓
- A rectangle (non-square): more than one angle of symmetry (180°, 360°) ✓
- An irregular shape/scalene triangle: only 360° — not more than one ✗

Note: Students should examine each figure shown and apply the rotation test.

Given: Six figures.

Concept: The order of rotational symmetry of a figure is the number of times the figure looks exactly the same during a complete rotation of $360°$ about its centre.
$$\text{Order} = \frac{360°}{\text{smallest angle of symmetry}}$$

Answer:

| Figure | Description | Smallest Angle | Order |
|---|---|---|---|
| 1 | 2-armed/S-shape | $180°$ | 2 |
| 2 | 3-armed/triangle | $120°$ | 3 |
| 3 | 4-armed/square | $90°$ | 4 |
| 4 | 5-armed/pentagon | $72°$ | 5 |
| 5 | 6-armed/hexagon | $60°$ | 6 |
| 6 | Irregular/asymmetric | $360°$ | 1 |

Formula used:
$$\text{Order of rotational symmetry} = \frac{360°}{\text{smallest angle of symmetry}}$$

Note: Order 1 means the figure has no rotational symmetry (it only maps to itself after a full $360°$ rotation).

Given: A circle divided into equal sectors (appears to be 12 sectors based on typical textbook figures).

Concept: The angles of symmetry of a coloured circle depend on how the colours repeat. If the pattern repeats every $k$ sectors out of $n$ total sectors, the smallest angle of symmetry is $\dfrac{360° \times k}{n}$.

Answer:

Part (i) — 3 angles of symmetry ($120°, 240°, 360°$):
Colour the sectors so that the pattern repeats every $\dfrac{1}{3}$ of the circle.
For a 12-sector circle: colour sectors 1, 2, 3, 4 in a pattern, then repeat the same pattern for sectors 5–8 and 9–12.
This gives rotational symmetry of order 3, with angles of symmetry: $120°, 240°, 360°$.

Part (ii) — 4 angles of symmetry ($90°, 180°, 270°, 360°$):
Colour the sectors so that the pattern repeats every $\dfrac{1}{4}$ of the circle.
For a 12-sector circle: colour sectors 1, 2, 3 in a pattern, then repeat for sectors 4–6, 7–9, and 10–12.
This gives rotational symmetry of order 4, with angles of symmetry: $90°, 180°, 270°, 360°$.

Part (iii) — Possible numbers of angles of symmetry:
For a circle with $n$ sectors, the possible orders of rotational symmetry are all divisors of $n$.
For $n = 12$: divisors are $1, 2, 3, 4, 6, 12$.
So the possible numbers of angles of symmetry are: 1, 2, 3, 4, 6, and 12.
(Order 1 means only $360°$, so 1 angle; order 2 gives 2 angles; etc.)

Given: Need figures (not circle or square) with both reflection symmetry (lines of symmetry) and rotational symmetry (angles of symmetry other than just 360°).

Concept: A figure has both types of symmetry if it has at least one line of symmetry AND at least one angle of symmetry strictly between 0° and 360°.

Answer:

Figure 1 — Equilateral Triangle:
- Reflection symmetry: 3 lines of symmetry (from each vertex to midpoint of opposite side).
- Rotational symmetry: Angles of symmetry are $120°, 240°, 360°$ (order 3).

Figure 2 — Regular Hexagon:
- Reflection symmetry: 6 lines of symmetry.
- Rotational symmetry: Angles of symmetry are $60°, 120°, 180°, 240°, 300°, 360°$ (order 6).

Other examples: Regular pentagon (5 lines, order 5), rectangle (2 lines, order 2), rhombus (2 lines, order 2).

Given: Various conditions on triangles and quadrilaterals.

Answer:

Part (a) — Triangle with at least 2 lines of symmetry and at least 2 angles of symmetry:
Draw an equilateral triangle.
- It has 3 lines of symmetry.
- It has angles of symmetry: $120°, 240°, 360°$ — so 3 angles of symmetry (at least 2 ✓).
*(Note: No triangle can have exactly 2 lines of symmetry. If it has 2, it must have 3, making it equilateral.)*

Part (b) — Triangle with only one line of symmetry but no rotational symmetry:
Draw an isosceles triangle (two equal sides, but not equilateral).
- It has 1 line of symmetry (perpendicular bisector of the base).
- It has no rotational symmetry (the only angle of symmetry is $360°$). ✓

Part (c) — Quadrilateral with rotational symmetry but no reflection symmetry:
Draw a parallelogram (that is not a rectangle or rhombus).
- It has rotational symmetry of order 2 (angle of symmetry: $180°$). ✓
- It has no lines of symmetry. ✓

Part (d) — Quadrilateral with reflection symmetry but no rotational symmetry:
Draw a kite (two pairs of adjacent equal sides, but not a rhombus).
- It has 1 line of symmetry (along the main diagonal). ✓
- It has no rotational symmetry (cannot be rotated by any angle less than $360°$ to look the same). ✓

Given: Smallest angle of symmetry $= 60°$.

Concept: All angles of symmetry are multiples of the smallest angle of symmetry, up to and including $360°$.

Working:
$$\text{Order of rotational symmetry} = \frac{360°}{60°} = 6$$

The angles of symmetry are all multiples of $60°$ up to $360°$:
$$60°, \ 120°, \ 180°, \ 240°, \ 300°, \ 360°$$

Answer: The other angles of symmetry are $\mathbf{120°, 180°, 240°, 300°, 360°}$.

Given: $60°$ is an angle of symmetry. There are two angles of symmetry less than $60°$.

Concept: All angles of symmetry are multiples of the smallest angle. If there are two angles less than $60°$, they are the 1st and 2nd multiples of the smallest angle, and $60°$ is the 3rd multiple.

Working:
Let the smallest angle of symmetry $= x$.

Angles less than $60°$: $x$ and $2x$ (two angles).
The next angle is $3x = 60°$.

$$3x = 60°$$
$$x = \frac{60°}{3} = 20°$$

Verification: Angles of symmetry: $20°, 40°, 60°, 80°, \ldots, 360°$ — yes, there are two angles ($20°$ and $40°$) less than $60°$. ✓

Answer: The smallest angle of symmetry is $\mathbf{20°}$.

Given: Checking if figures can have specific smallest angles of symmetry.

Concept: A figure can have a smallest angle of symmetry of $x°$ if and only if $x$ is a divisor of $360°$, i.e., $\dfrac{360°}{x}$ is a whole number (positive integer). This is because the order of rotational symmetry must be a whole number.

Part (a) — Smallest angle $= 45°$:
$$\frac{360°}{45°} = 8$$
$8$ is a whole number. ✓

Yes, a figure can have $45°$ as its smallest angle of symmetry. It would have order 8 rotational symmetry.
Example: A regular octagon has smallest angle of symmetry $= 45°$.

Part (b) — Smallest angle $= 17°$:
$$\frac{360°}{17°} = 21.17\ldots$$
This is not a whole number. ✗

No, a figure cannot have $17°$ as its smallest angle of symmetry, because $17$ does not divide $360$ evenly.

Answer: (a) Yes, possible. (b) No, not possible.

Given: Picture of the new Parliament Building in Delhi (triangular/triagonal shape).

Concept: Reflection symmetry requires lines along which the figure is a mirror image of itself. Rotational symmetry requires the figure to look the same after rotation by certain angles.

Answer:

Part (a) — Reflection symmetry:
The outer boundary of the new Parliament Building is roughly triangular (it has a triangular footprint).
A regular triangle has 3 lines of symmetry.
However, the building's outer boundary appears to have 1 line of symmetry — a vertical line through the apex of the triangle to the midpoint of the base (if it is an isosceles triangle shape).

Lines of symmetry: 1 (the vertical axis of symmetry through the centre).

Part (b) — Rotational symmetry:
If the building has a triangular shape with 3-fold symmetry (equilateral triangle), it would have rotational symmetry with angles: $120°, 240°, 360°$.

If it is only isosceles (not equilateral), it has no rotational symmetry (other than $360°$).

Based on the actual building: The Parliament Building has a roughly equilateral triangular outer boundary, giving it 3-fold rotational symmetry with angles of symmetry: $\mathbf{120°, 240°, 360°}$.

Given: Regular polygons: equilateral triangle, square, regular pentagon, regular hexagon, ...

Concept: A regular polygon with $n$ sides has exactly $n$ lines of symmetry.

Working:
| Regular Polygon | Number of sides ($n$) | Lines of Symmetry |
|---|---|---|
| Equilateral Triangle | 3 | 3 |
| Square | 4 | 4 |
| Regular Pentagon | 5 | 5 |
| Regular Hexagon | 6 | 6 |
| Regular Heptagon | 7 | 7 |
| $\vdots$ | $\vdots$ | $\vdots$ |

Number sequence obtained: $3, 4, 5, 6, 7, 8, \ldots$

This is the sequence of natural numbers starting from 3, i.e., $n$ for $n = 3, 4, 5, 6, \ldots$

Given: Regular polygons: equilateral triangle, square, regular pentagon, regular hexagon, ...

Concept: A regular polygon with $n$ sides has rotational symmetry of order $n$, meaning it has exactly $n$ angles of symmetry (including $360°$).

Working:
| Regular Polygon | Sides ($n$) | Angles of Symmetry |
|---|---|---|
| Equilateral Triangle | 3 | 3 |
| Square | 4 | 4 |
| Regular Pentagon | 5 | 5 |
| Regular Hexagon | 6 | 6 |
| $\vdots$ | $\vdots$ | $\vdots$ |

Number sequence obtained: $3, 4, 5, 6, 7, 8, \ldots$

This is the same sequence as the lines of symmetry: natural numbers starting from 3.

Note: For a regular $n$-gon, the number of lines of symmetry equals the number of angles of symmetry, both equal to $n$.

Given: The Koch Snowflake sequence (fractal snowflake shapes).

Concept: The Koch Snowflake is built starting from an equilateral triangle. At each stage, the symmetry of the equilateral triangle is preserved.

Answer:

Lines of symmetry:
At every stage of the Koch Snowflake construction, the figure retains the symmetry of the original equilateral triangle.
$$\text{Lines of symmetry} = \mathbf{3}$$
(The same 3 lines of symmetry as the equilateral triangle, at every stage.)

Angles of symmetry:
Similarly, the rotational symmetry of order 3 is preserved at every stage.
$$\text{Angles of symmetry} = 120°, 240°, 360° \quad (\text{i.e., } \mathbf{3} \text{ angles of symmetry})$$

Answer: Each shape in the Koch Snowflake sequence has 3 lines of symmetry and 3 angles of symmetry.

Given: The Ashoka Chakra (the wheel on the Indian national flag).

Concept: The Ashoka Chakra is a circle with 24 spokes equally spaced.

Working:

Lines of symmetry:
Each spoke and the gap between opposite spokes forms a line of symmetry. With 24 equally spaced spokes:
$$\text{Lines of symmetry} = 24$$
(Each line passes through a spoke and the diametrically opposite spoke, or through the midpoints between spokes.)

Angles of symmetry:
The smallest angle of symmetry:
$$\text{Smallest angle} = \frac{360°}{24} = 15°$$

All angles of symmetry are multiples of $15°$:
$$15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°, 195°, 210°, 225°, 240°, 255°, 270°, 285°, 300°, 315°, 330°, 345°, 360°$$

$$\text{Number of angles of symmetry} = 24$$

Answer: The Ashoka Chakra has $\mathbf{24}$ lines of symmetry and $\mathbf{24}$ angles of symmetry (multiples of $15°$ from $15°$ to $360°$).