Chapter 9: Symmetry

Step 1  Recall the idea of a line of symmetry
A line of symmetry (also called an axis of symmetry) for a figure is a straight line which divides the figure into two identical halves. If we fold the figure along that line, the two halves fit exactly one over the other.

Step 2  Look again at the figures printed at the very beginning of the chapter
(Your textbook shows a collection of familiar shapes – for example a leaf, a butterfly, a mask/face, a star-fish, the front view of a building, a wheel, etc. – together with a sketch of a fluffy cloud.) One by one, check whether each shape can be folded so that its two parts coincide.

Step 3  Conclusion
Every figure shown at the start of the chapter, except the cloud, possesses at least one line of symmetry. The sketch of the cloud is an uneven, free-hand outline; because its left and right (or top and bottom) parts cannot be matched exactly in any way, it has no line of symmetry.

Concept recalled
A line of symmetry (also called an axis of symmetry) is a straight line about which a figure can be folded so that the two halves coincide exactly. While looking for such a line we can imagine (or actually try by paper–folding) placing one half of the figure over the other; if they match perfectly, the crease gives a line of symmetry.

Below every part-figure is discussed exactly the way a Class 6 student can check by folding, tracing or using a mirror. Each time we state how many lines of symmetry exist and name (or sketch) them.

Note for drawing during study – in your notebook reproduce the figure, draw the claimed axis/axes with a ruler and then fold along that line to verify.

(a) Figure (a) – an isosceles triangle having the two equal sides slanting upwards and meeting at a top vertex.

• Place the base horizontally. Try folding along the vertical line through the top vertex and the mid-point of the base.
• The left half comes exactly over the right half → the fold is a line of symmetry.
• Any other fold (say horizontal or slant) leaves the halves unmatched.

Hence figure (a) has 1 line of symmetry – the vertical line through the top vertex and mid-point of the base.

(b) Figure (b) – an arrow pointing right, drawn so that its upper and lower edges are mirror images.

• Keep the arrow pointing right. Fold along the horizontal line passing through the middle of the arrow shaft (equal distance from its top and bottom edges).
• The top half covers the bottom half perfectly → horizontal line is a line of symmetry.
• The vertical line through the centre fails (the arrow head does not match its tail); diagonals also fail.

Therefore figure (b) possesses exactly 1 line of symmetry – the horizontal middle line.

(c) Figure (c) – a cross-shaped design whose left half looks like the right half **and** whose upper half looks like the lower half.

• Fold vertically: halves coincide → vertical line is a symmetry.
• Fold horizontally: halves also coincide → horizontal line is another symmetry.
• Folding along either diagonal does not work because the arms of the cross differ in length along the diagonal direction.

So figure (c) has 2 lines of symmetry – one vertical and one horizontal.

(d) Figure (d) – a perfect square.

• A square is well-known to be the most symmetric quadrilateral.
• Vertical and horizontal diameters are axes of symmetry.
• Both diagonals are also axes (because opposite vertices and side lengths match under this fold).

Hence the square has 4 lines of symmetry – vertical, horizontal and the two diagonals.

(e) Figure (e) – a general scalene triangle (all sides different).

• Try every conceivable fold – no single line can bring all three vertices into matching positions because side lengths and angles are all unequal.

Thus figure (e) has no line of symmetry.

(f) Figure (f) – the English capital letter ‘A’ written in block style with equal feet.

• Fold along the vertical mid-line that passes through the apex and the mid-point of the bar joining the two legs. The left leg overlaps the right leg and the bar matches itself.
• Any other fold fails.

Therefore figure (f) has 1 line of symmetry – the vertical axis.

(g) Figure (g) – the block capital letter ‘B’ drawn with two equal semicircular bulges.

• Keeping the letter upright, fold along the horizontal line that passes through the middle of the vertical backbone; the upper bulge coincides with the lower bulge.
• Vertical and diagonal folds fail.

Thus figure (g) has 1 line of symmetry – the horizontal middle line.

(h) Figure (h) – a circle.

• Choose any diameter, fold the circle along that diameter – the halves coincide. Because one can draw infinitely many diameters, a circle has infinitely many axes of symmetry.

So figure (h) has infinitely many lines of symmetry – every diameter is an axis.

Summary table

All required lines of symmetry have been found and justified.

Step 1 : Recall what it means to fold a shape so that the two halves overlap
When we fold a paper model of a figure, if every point on one part of the figure exactly covers a point on the other part, the crease made by the fold is called a line of symmetry (or axis of symmetry).

Step 2 : Draw and name the obvious fold already shown in the book
Suppose the book has already shown the diagonal from the top-left corner to the bottom-right corner. That diagonal is one line of symmetry because folding along it makes the two triangular halves coincide.

Step 3 : Look for all possible creases that pass through the centre
Because a square is perfectly regular, any line that passes through its centre $$O$$ and produces two mirror-image parts will be a line of symmetry.

• Joins the mid-points of opposite sides: there are two such lines.
    • The line joining the mid-points of the left and right sides (a vertical line).
    • The line joining the mid-points of the top and bottom sides (a horizontal line).
• Joins opposite vertices: there are two such lines.
    • The diagonal from the top-left corner to the bottom-right corner.
    • The diagonal from the top-right corner to the bottom-left corner.

Each of these four lines goes through the centre of the square and divides it into two congruent (identical) parts. Therefore folding along any of these lines makes the two halves overlap exactly.

Step 4 : Count the lines of symmetry
Since no other straight line through the centre leaves the square unchanged except the four listed above, the total number of lines of symmetry of a square is
\[\text{Number of lines of symmetry of a square} = 4\]

Conclusion
Yes. Besides the diagonal shown earlier, you can also fold along the other diagonal, the vertical centre line and the horizontal centre line. Altogether, a square has four lines of symmetry.

Step 1 Recap of idea of line of symmetry
A straight line is called a line of symmetry of a figure if, when the figure is folded (or reflected) about that line, every point of the figure falls exactly on a matching point of the figure; that is, the two separated parts overlap completely.

Step 2 Draw and inspect a non‑square rectangle
Take a rectangle $$ABCD$$ with length $$AB = CD = l$$ and breadth $$BC = AD = b$$ such that $$l \neq b$$. Draw one of its diagonals, say $$AC$$.

Visual test If we try to imagine reflecting (or folding) the rectangle about line $$AC$$, the long side $$AB$$ would have to land on the short side $$AD$$. Because $$AB$$ and $$AD$$ are of different lengths (they are $$l$$ and $$b$$ respectively), they cannot coincide. Hence the whole figure will not match up.

Step 3 Concrete paper-folding test

1. Cut a neat rectangular sheet whose length and breadth are unequal.
2. Draw one diagonal and fold the paper exactly along this diagonal.
3. Look at the two halves formed by the fold.

Observation: The two triangular halves overlap only partly; some portion of one triangle sticks out because one side of each triangle is $$l$$ while the corresponding side of the other is $$b$$. The halves therefore do not cover each other completely.

Step 4 Conclusion
Because folding about either diagonal does not make the two portions coincide perfectly, a diagonal in a rectangle whose adjacent sides are unequal is not a line of symmetry. Only the two lines that join the mid-points of opposite sides (the horizontal and the vertical mid-lines) are lines of symmetry for such a rectangle. In a square, where $$l = b$$, the mismatch disappears, so each diagonal does become a line of symmetry, but not in an ordinary rectangle.

Given: A square $$ABCD$$ with the usual order of vertices (travelling clockwise):
A – top left, B – top right, C – bottom right, D – bottom left.

Two different axes of symmetry are to be used:

1. the diagonal $$AC$$, and
2. the horizontal line passing through the mid-points of $$AB$$ and $$CD$$ (call this line $$ ext{l}_h$$).

You may draw two separate sketches of the square and mark the required axes. In each case place your compass or a folding strip exactly on the given line to check which points coincide.

1. Reflection in the diagonal $$AC$$

• The diagonal $$AC$$ itself is the mirror. Any point on the mirror stays where it is.
  Therefore $$A \rightarrow A$$ and $$C \rightarrow C$$.
• Point $$B$$ is the same perpendicular distance from $$AC$$ as point $$D$$ but lies on the opposite side of the diagonal. Hence they interchange their positions:
  $$B \rightarrow D\;\;\text{and}\;\;D \rightarrow B$$.

So the complete mapping is

$$A \mapsto A,\;B \mapsto D,\;C \mapsto C,\;D \mapsto B.$$

2. Reflection in the horizontal axis $$\text{l}_h$$

• This horizontal line cuts the square into a top half and a bottom half. Points lying directly above the line go to the corresponding points directly below it, and vice-versa.
• Vertex $$A$$ (top left) is exactly above $$D$$ (bottom left), hence
  $$A \rightarrow D\;\;\text{and}\;\;D \rightarrow A$$.
• Vertex $$B$$ (top right) is exactly above $$C$$ (bottom right), hence
      $$B \rightarrow C\;\;\text{and}\;\;C \rightarrow B$$.

The mapping for this reflection is therefore

$$A \mapsto D,\;B \mapsto C,\;C \mapsto B,\;D \mapsto A.$$

How to remember: Every reflection keeps points lying on the mirror fixed and sends all other points to the unique points that are the same perpendicular distance on the opposite side of the mirror.

Key Idea :  When a sheet is folded, every point of the sheet sits exactly over the point that is its reflection in the fold line. After a hole is punched through all the layers and the paper is unfolded, the pattern of holes is therefore symmetric about the fold line. So, to discover the line of fold we only have to look for an axis of symmetry in the completed (unfolded) picture.

We examine each figure one by one.

1. Figure (a)
   The two identical holes lie at equal distances on opposite sides of an imaginary vertical central line. Folding the paper about this vertical mid-line would bring the left hole exactly over the right one. Hence the fold line is the vertical line that cuts the square into two equal left–right halves.
   $$\text{Fold line for (a)}: \;\text{Vertical middle line}$$
2. Figure (b)
   Here the pair of holes is placed symmetrically with respect to a diagonal that runs from the top-left corner to the bottom-right corner. If the sheet is folded about this diagonal, each hole covers the other. Therefore the diagonal through those two opposite corners is the required line of fold.
   $$\text{Fold line for (b)}: \;\text{Diagonal } (\text{top-left }\rightarrow\text{ bottom-right})$$
3. Figure (c)
   Both holes are mirror images of each other across a horizontal mid-line. Folding the square about the horizontal centre line makes the upper part coincide with the lower part, matching the two holes. Hence that horizontal line is the fold line.
   $$\text{Fold line for (c)}: \;\text{Horizontal middle line}$$
4. Figure (d)

In the unfolded sheet we now see four identical holes forming the vertices of a smaller square that is centred on the middle of the big square. The whole pattern is symmetric about both the vertical and the horizontal centre lines. Yet the statement says only one hole was actually punched. This is possible only if the square was folded twice before the punching, so that all four layers lay one over the other.

• Step 1 : Fold the original square once about the vertical mid-line. The left half covers the right half.
• Step 2 : Fold this half-square again about the horizontal mid-line. The top now covers the bottom, giving one quarter of the original sheet, four layers thick.
• Step 3 : Punch a single hole through the four layers at the position marked in the drawing.
• Step 4 : Unfold the paper. Each unfolding doubles the number of holes, so the one hole becomes two after the first unfolding and four after the second, producing exactly the arrangement shown.

Thus, for Figure (d) the sheet was first folded along one centre line and then along the other centre line at right angles, giving a quarter-sized layered sheet before the single hole was made.

Summarising :

Understanding the problem
When a sheet of paper is said to have a line of symmetry, it means that, on folding the paper along this line, every point on one side falls exactly on a matching point on the other side. If a hole is punched on one side of the line, then, after unfolding, we must see on the other side another hole that is the mirror image of the original one.

The textbook question shows a dotted line (or more than one dotted line) and one or more punched holes. We have to mark the points where the remaining holes will appear when the paper is unfolded.

Step–by–step method

1. Draw the line(s) of symmetry clearly.
     Mark the given dotted line as the mirror line.
2. Take one given hole at a time.
     Measure its perpendicular (shortest) distance to the mirror line.
3. Locate the image.
     On the other side of the mirror line, mark a point at the same perpendicular distance and on the same straight line that passes through the original hole and is perpendicular to the mirror line.
4. Repeat the above two steps for every given hole.
5. If two lines of symmetry are shown, first reflect in one line, then reflect each of the resulting holes in the second line. (Every reflection doubles the number of holes.)
6. Finally, count and label all the holes. The diagram should now look identical on both (or all) sides of every line of symmetry drawn.

Why the construction works

• Reflection keeps the distances of points from the mirror line intact but reverses the side (left becomes right or top becomes bottom, etc.).
• Hence every original point and its image are collinear with, and equidistant from, the mirror line — that is the defining property of symmetry.

Illustration to draw (describe in words)
Redraw the given figure. For each shown hole, draw a short perpendicular to the dotted line. Using a ruler, mark an equal-length perpendicular on the opposite side. Show the new point as a similar tiny disc (a punched hole). When two mirror lines intersect (for example, a vertical and a horizontal), first reflect as described above in one line, then take every hole (old and new) and repeat the rule with respect to the second mirror line.

After completing all reflections, you will have the full set of holes that make the figure symmetric.

Question : Exercise 9.1 (NCERT Class 6, Chapter 9 – Symmetry)
Find the number of lines of symmetry for each of the following shapes:
(i) an equilateral triangle  (ii) an isosceles triangle  (iii) a scalene triangle  (iv) a square  (v) a rectangle  (vi) a rhombus  (vii) a parallelogram  (viii) a regular hexagon  (ix) a circle.

Solution :

1. Equilateral triangle
   All three sides are equal and all three angles are $$60^{\circ}$$.
   Each altitude also bisects the opposite side and the vertex angle, therefore acts as an axis of symmetry.
   Thus there are $$3$$ lines of symmetry – one through every vertex and the midpoint of the opposite side.

2. Isosceles triangle
   Two sides are equal. The perpendicular from the vertex joining the equal sides to the base bisects the base and the vertex angle.
   This single line is the only axis of symmetry, so the number is $$1$$.

3. Scalene triangle
   No two sides (or angles) are equal, so no fold can make one half coincide exactly with the other.
   Hence the number of lines of symmetry is $$0$$.

4. Square
   All four sides are equal and all angles are right angles.
   The two diagonals and the two lines joining the mid-points of opposite sides are axes of symmetry.
   Total lines of symmetry $$= 4$$.

5. Rectangle (not a square)
   Opposite sides are equal; the diagonals are not axes because they do not map every vertex onto another vertex of equal distance.
   The only axes are the two lines joining the mid-points of opposite sides (one vertical, one horizontal).
   Hence, $$2$$ lines of symmetry.

6. Rhombus (not a square)
   All sides are equal but angles are not right angles.
   The two diagonals bisect each other at right angles and also bisect the interior angles; they are the only axes.
   Therefore, $$2$$ lines of symmetry.

7. Parallelogram (not a rectangle/rhombus)
   Opposite sides are equal and parallel but neither the sides nor the angles come in pairs that allow a fold to match the figure with itself.
   So it has $$0$$ lines of symmetry.

8. Regular hexagon
   For a regular $$n$$-gon the number of symmetry lines equals $$n$$.
   Therefore a regular hexagon has $$6$$ lines of symmetry:
   − three lines joining opposite vertices and
   − three lines joining the mid-points of opposite sides.

9. Circle
   Every diameter divides the circle into two equal halves.
   Because there are infinitely many diameters, a circle has infinitely many (uncountably many) lines of symmetry.

Conclusion : The required numbers are summarised below.

Question : Trace the given square and draw all its lines of symmetry, if any.

Solution :

1. Identify the figure – The shape is a square: all four sides are equal and each interior angle is $$90^{\circ}$$.

2. Recall the idea of symmetry – A line of symmetry (or mirror line) is a straight line such that folding the figure along it makes the two halves coincide exactly.

3. Find possible mirror lines

   • Vertical line $$L_1$$: Join the mid-points of the left and right sides. Folding along $$L_1$$ matches the left half with the right half, so $$L_1$$ is a line of symmetry.

   • Horizontal line $$L_2$$: Join the mid-points of the top and bottom sides. Folding along $$L_2$$ makes the top half coincide with the bottom half, hence $$L_2$$ is a line of symmetry.

   • First diagonal $$L_3$$: Join the top-left vertex to the bottom-right vertex. Folding along $$L_3$$ brings one triangular half exactly over the other, so $$L_3$$ is a line of symmetry.

   • Second diagonal $$L_4$$: Join the top-right vertex to the bottom-left vertex. Folding along $$L_4$$ also gives matching halves, so $$L_4$$ is a line of symmetry.

4. Verify completeness – Any other straight line through the square fails the folding test. Therefore, the square has exactly the four lines $$L_1, L_2, L_3, L_4$$.

5. How to draw on the tracing

   1. Copy the square on tracing paper.
   2. Draw the vertical line through the mid-points of left and right sides.
   3. Draw the horizontal line through the mid-points of top and bottom sides.
   4. Draw the two diagonals joining opposite vertices.

   The four drawn lines represent all the lines of symmetry of the square.

Result: The square possesses $$4$$ lines of symmetry.

Step 1 : Observe the figure carefully
The centre of the kolam is the point where all its curved arms seem to meet. Look for parts that repeat or appear to be mirror images of each other.

Step 2 : Test a vertical fold
Imagine (or actually draw) a straight line passing through the centre from top to bottom. If you fold the paper along this imagined line, the left half of the kolam exactly covers the right half. Hence the vertical centre line is a line of symmetry.

Step 3 : Test a horizontal fold
Next draw a straight line through the centre from left to right. On folding about this line, the upper half coincides with the lower half. Therefore the horizontal centre line is also a line of symmetry.

Step 4 : Test the two diagonals
Now draw a slanting line through the centre joining the top-left part of the kolam to the bottom-right part. Try folding the figure along this line: every point on the part above the line falls exactly on a corresponding point below it. So this diagonal is a line of symmetry.
Repeat the check for the other diagonal (joining the top-right part to the bottom-left part). The kolam again matches up perfectly, so this second diagonal is also a line of symmetry.

Step 5 : List all lines of symmetry
The kolam can be folded onto itself in four different ways:

• one vertical line through the centre,
• one horizontal line through the centre,
• one diagonal (top-left to bottom-right),
• the other diagonal (top-right to bottom-left).

Hence the kolam possesses exactly four lines of symmetry.

Preliminary note about dot grids
A dot grid is just a sheet of equally-spaced points (like a square sheet of graph paper but with dots instead of lines). Each adjacent pair of dots, either horizontally or vertically, is one unit apart. When we are asked to “draw two more lines” we must join pairs of dots with straight segments. All counting of distances below is therefore done in whole units measured from dot to dot.

The three sub-figures supplied in the textbook are reproduced first. For each of them we then add exactly two new straight segments so that the completed shape becomes bilaterally (mirror) symmetric. In every case the required line of symmetry is shown as a dotted line, and the two extra solid segments are described step-by-step.

(a) First figure

• Copy the given ‘┐’ shaped outline on the dot grid. The open end of the shape faces the left.
• Count the width of the open mouth: from the last drawn vertical segment go four dots to the left. Through the middle of this gap draw a dotted vertical line; this will be the intended line of symmetry.
• Now add the two solid segments:
  1. Join the upper end of the existing right-hand vertical to the dot that is symmetric with the lower end (one unit above the bottom horizontal). That gives a short horizontal segment on the left.
  2. Join the lower end of the existing right-hand vertical straight down to meet the bottom horizontal. That gives a matching vertical on the left.
  After drawing these two lines the outline on the left is the mirror image of the outline on the right, so the whole figure now has a vertical line of symmetry.

(b) Second figure

• Copy the given ‘Γ’ shaped outline (looks like an inverted L) so that the open side faces upward.
• Measure the height of the opening: it is three dots. Draw a dotted horizontal line that passes through the middle dot of this gap; this will be the desired line of symmetry.
• Add the two needed solid segments:
  1. From the right-hand end of the existing top horizontal, draw a horizontal segment of equal length to the left. This closes the top exactly in the middle.
  2. From the upper end of the existing left-hand vertical, drop a vertical segment of equal length downward. This closes the left side exactly in the middle.
  Folding the paper about the dotted horizontal now makes the two halves coincide; hence the shape is symmetric about that horizontal line.

(c) Third figure

• The supplied outline looks like a right angle open towards the south-west.
• Count the number of dots that separate the two arms of the right angle: it is two in each direction. Draw a dotted diagonal line that slopes downwards from left to right and cuts the corner exactly through those middle dots. That diagonal will serve as the line of symmetry.
• Finally add the two solid segments:
  1. Join the lower end of the right-hand arm to the dot that is symmetric across the diagonal (two units up and two units left). That gives the first closing side.
  2. Join the left-hand end of the upper arm to the corresponding symmetric dot (two units right and two units down). That gives the second closing side.
  Because each new segment has been drawn by reflecting an existing side in the chosen diagonal, the completed quadrilateral now has the required diagonal line of symmetry.

Why only two extra lines are sufficient each time
In every sub-question the given outline already supplies one half of the intended symmetric figure. The new two lines are exactly the images (under reflection) of two existing sides, so together they supply the missing half. Once the outline is closed in this way, reflection in the indicated axis swaps every side with its partner, proving that a line of symmetry really exists.

Step 1 : Recall the idea of rotational symmetry
A figure has rotational symmetry if, when we turn (rotate) it about a fixed point (here, the marked dot), it fits exactly on to itself before a full $$360^{\circ}$$ turn is completed. The smallest positive angle for which this happens is called the angle of rotational symmetry, and the number of such exact fittings in one full turn is the order of rotational symmetry:

• Order $$=\dfrac{360^{\circ}}{\text{smallest positive angle}}.$$

Step 2 : Identify the centre of rotation
The black dot drawn at the middle of the strip shows the centre of rotation. All rotations will be performed about this point.

Step 3 : Test the usual factors of $$360^{\circ}$$
A convenient way is to test the divisors of $$360^{\circ}$$ (that is, $$90^{\circ},\;120^{\circ},\;180^{\circ},\;\ldots$$) and check whether the strip coincides with itself.

1. Rotation through $$90^{\circ}$$.
     Because the strip is a slanted, parallelogram-shaped band, a quarter-turn would make its long edges lie where its short edges were. The outline therefore does not match the original position. So $$90^{\circ}$$ is not a symmetry angle.
2. Rotation through $$120^{\circ}$$ or $$60^{\circ}$$.
     These angles leave threefold or sixfold symmetry, typical of equilateral-triangle or hexagonal figures, which the strip clearly is not. Hence they also fail.
3. Rotation through $$180^{\circ}$$.
     Turning the strip half-way round brings each long edge to the place of the opposite long edge and similarly for the short edges. The dot at the centre stays put. The whole figure therefore fits exactly on to itself. Thus $$180^{\circ}$$ is a symmetry angle.
4. Rotation through $$270^{\circ}$$.
     A three-quarter turn gives the same result as the earlier $$90^{\circ}$$ test—no match, so no symmetry.
5. Rotation through $$360^{\circ}$$.
     Every shape coincides with itself after a full turn. Although this is always true, we usually do not count it when we talk about “symmetry other than the trivial one”.

Step 4 : State the order and angle(s)
The smallest positive angle that works is $$180^{\circ}$$, so

• angle of rotational symmetry  =  $$180^{\circ}$$,
• order of rotational symmetry  =  $$\dfrac{360^{\circ}}{180^{\circ}} = 2$$.

Answer
Hence the strip possesses rotational symmetry through an angle of $$180^{\circ}$$ (and, of course, also through $$360^{\circ}$$).

Background idea

When a figure looks exactly the same after a rotation about its centre, the amount of turn is called an angle of rotational symmetry. If one complete turn is divided into n equal parts, the figure will repeat its appearance n times, giving exactly n angles of symmetry.

(a) A figure with exactly five angles of symmetry

1. Decide the number of arms.
   The hint asks us to use 5 radial arms. Equal spacing will make the arm-pattern repeat 5 times in one full turn.
2. Angle between adjacent arms.
   A full turn is $$360^{\circ}$$. Five equal parts give \[\dfrac{360^{\circ}}{5}=72^{\circ}\] So each neighbouring pair of arms must be separated by $$72^{\circ}$$.
3. How to draw.
   • Draw a point O as centre.
   • Draw a ray from O in any direction – that is arm 1.
   • With a protractor mark $$72^{\circ}$$ from that ray and draw arm 2.
   • Keep marking $$72^{\circ}$$ successively and draw arm 3, arm 4 and arm 5.
   • All five arms should be of equal length for neatness.
   This star-like figure repeats every $$72^{\circ}$$, so it has five angles of rotational symmetry: $$72^{\circ},\;144^{\circ},\;216^{\circ},\;288^{\circ},\;360^{\circ}.$$

(b) A figure with exactly six angles of symmetry

1. Choose 6 arms. A 6-armed radial figure will repeat 6 times in one turn.
2. Angle between arms. \[\dfrac{360^{\circ}}{6}=60^{\circ}\] Each pair of neighbouring arms must be $$60^{\circ}$$ apart.
3. Drawing steps.
   • Mark centre O.
   • Draw the first ray, then successively mark $$60^{\circ}$$ five times more to obtain six equal rays.
   The figure matches itself after every $$60^{\circ}$$ turn, giving the six required symmetry angles: $$60^{\circ},\;120^{\circ},\;180^{\circ},\;240^{\circ},\;300^{\circ},\;360^{\circ}.$$

Why there are no more (or fewer) angles

Because the arms are equal and equally spaced, the smallest turn that places every arm exactly on the next one is $$\dfrac{360^{\circ}}{n}$$, where n is the number of arms. All other matching angles are whole-number multiples of this basic turn, so the total number of different angles is exactly n – no more, no less.

Step 1 – Recall the meaning of "angle of symmetry" for a radial (rotational) design
For any figure that can be rotated about a fixed centre:

• If the figure looks exactly the same after turning through a certain angle, that turn is called an angle of rotational symmetry.
• The smallest positive such turn is called the smallest angle of symmetry.
• The number of different positions in which the figure matches itself during a full turn of $$360^{\circ}$$ is called the order of rotational symmetry.

Step 2 – Connect the order with the smallest angle
If a shape has order $$n$$, then one full turn ( $$360^{\circ}$$ ) is cut into $$n$$ equal parts. Therefore

Step 3 – Insert the given order
The question says the figure has exactly 7 angles of symmetry, so its order is $$n=7$$.

Using (1):
$$\text{smallest angle}=\frac{360^{\circ}}{7}$$

Step 4 – Write the answer as a fraction and then as a mixed number

• Fractional form: $$\frac{360}{7}^{\circ}$$
• Long division (or using multiplication facts) gives
  $$360\div7=51\text{ remainder }3$$
• Hence
  $$\frac{360}{7}=51+\frac{3}{7}=51\frac{3}{7}$$

Step 5 – Conclude about whole‑number status
Because $$\frac{360}{7}$$ is not a whole number, the angle is expressed as the mixed fraction $$51\frac{3}{7}^{\circ}$$.

Idea to be used

• If, during one complete turn of 360 °, a figure fits exactly on to itself n times, we say it has rotational symmetry of order $$n$$.
• The smallest turning needed to bring the figure back to its original position is therefore $$\theta = \dfrac{360^\circ}{n}$$.
• Because the figure repeats itself after every such turn, the full list of angles that work is $$\theta,\;2\theta,\;3\theta,\;\ldots ,\;n\theta=360^\circ.$$

The four given diagrams (as printed in the NCERT textbook) are:

1. An equilateral triangle with the point $$\bullet$$ at its centre.
2. A non‑square rectangle with the point $$\bullet$$ at its centre.
3. A square with the point $$\bullet$$ at its centre.
4. A regular hexagon with the point $$\bullet$$ at its centre.

We now work out each case one by one.

(i) Equilateral triangle

• The triangle fits on to itself 3 times in one full turn ⇒ order of rotation $$n = 3$$.
• Smallest angle $$\theta = \dfrac{360^\circ}{3} = 120^\circ.$$
• Hence the admissible angles are $$120^\circ,\;240^\circ,\;360^\circ.$$

(ii) Rectangle (length \(\neq\) breadth)

• The rectangle matches its original outline only twice in a full turn ⇒ order $$n = 2$$.
• Smallest angle $$\theta = \dfrac{360^\circ}{2} = 180^\circ.$$
• Therefore the angles are $$180^\circ,\;360^\circ.$$

(iii) Square

• Because all four sides are equal, the square fits on to itself 4 times ⇒ order $$n = 4$$.
• Smallest angle $$\theta = \dfrac{360^\circ}{4} = 90^\circ.$$
• This gives the angles $$90^\circ,\;180^\circ,\;270^\circ,\;360^\circ.$$

(iv) Regular hexagon

• A regular hexagon repeats itself 6 times in one full turn ⇒ order $$n = 6$$.
• Smallest angle $$\theta = \dfrac{360^\circ}{6} = 60^\circ.$$
• Hence the angles are $$60^\circ,\;120^\circ,\;180^\circ,\;240^\circ,\;300^\circ,\;360^\circ.$$

Conclusion

Step 1 : What is an angle of symmetry?
An angle of symmetry of a figure is an angle through which the figure can be rotated about its centre so that it falls exactly on itself. Every figure has $$360^{\circ}$$ (a full turn) as a trivial angle of symmetry. A figure has more than one angle of symmetry exactly when some rotation smaller than a full turn also brings it back to itself.

If the smallest angle of symmetry is $$\theta$$, the angles of symmetry are $$\theta, 2\theta, 3\theta, \ldots, 360^{\circ}$$, and the total number of them is $$\dfrac{360^{\circ}}{\theta}$$.

Step 2 : Inspect every drawing
The textbook gives ten drawings, marked (a) to (j). For each one, imagine pinning the figure at its centre and turning it through one full turn. Count the number of distinct positions in which it falls exactly on the original outline.

Step 3 : Pick out the figures with more than one angle of symmetry

• (c) Equilateral triangle — 3 angles
• (d) Square — 4 angles
• (e) Rectangle — 2 angles
• (g) Regular pentagon — 5 angles
• (i) Regular hexagon — 6 angles
• (j) Circle — infinitely many angles

The remaining drawings (a), (b), (f) and (h) are irregular: apart from the trivial full turn no rotation brings them back to themselves, so each has exactly one angle of symmetry.

Key idea — order of rotational symmetry
When a shape is turned about its centre through $$360^{\circ}$$, the number of different positions in which it looks exactly the same is called its order of rotational symmetry.
If a figure matches itself every $$\theta^{\circ}$$, then

\[\text{order}=\dfrac{360^{\circ}}{\theta^{\circ}}\]

Below, imagine tracing each figure on transparent paper, putting a pin through the centre and slowly turning the paper one full round. Count how many times the outline falls exactly on the original.

1. Figure (i) — Rectangle (not a square)
   A rectangle coincides with itself after a half-turn $$180^{\circ}$$ and again after the full turn $$360^{\circ}$$.
   Number of coincidences = 2, so the order is $$2$$.

2. Figure (ii) — Equilateral triangle
   The triangle fits on itself every $$120^{\circ}$$ (three equal turns in a full circle).
   \[\text{order}=\frac{360^{\circ}}{120^{\circ}}=3\]

3. Figure (iii) — Rhombus (not a square)
   Like the rectangle, a rhombus looks the same after a half-turn $$180^{\circ}$$ and after the full turn $$360^{\circ}$$ only.
   Order = $$2$$.

4. Figure (iv) — Square
   A square coincides with itself every quarter-turn $$90^{\circ}$$.
   \[\text{order}=\frac{360^{\circ}}{90^{\circ}}=4\]

5. Figure (v) — Regular hexagon
   The hexagon matches its outline every $$60^{\circ}$$.
   \[\text{order}=\frac{360^{\circ}}{60^{\circ}}=6\]

Therefore
(i) 2, (ii) 3, (iii) 2, (iv) 4, (v) 6.

Let a number of lines (or rays) are drawn starting from the same point O and they are equally spaced – this is exactly what happens when we join the centre of a regular figure (or a symmetric design) to its similar points.

1. One complete turn about O measures $$360^{\circ}$$.
2. Suppose there are n such equally spaced rays. Then all of them cut the full $$360^{\circ}$$ into n equal parts, so the size of the smallest central angle is \[\frac{360^{\circ}}{n}.\] 3. Any other angle we look at is formed by taking two of these rays. If we count how many of the equal little sectors lie between those two rays, say k of them, the obtained angle will be $$k\times\frac{360^{\circ}}{n}=\frac{k}{n}\times360^{\circ}.$$ Here k is a whole number (1, 2, 3, …) because we are counting complete small sectors.
4. Therefore every possible angle is a whole-number (integral) multiple of the smallest angle $$\frac{360^{\circ}}{n}$$.

Hence, yes – whenever the rays are equally spaced, every angle that appears must indeed be a multiple of the smallest angle. This is why, in all the cases you observed, the angles turned out to be such multiples, and it will always be so.

Key idea 1 : Angle (or angle of rotation) of symmetry
For rotational symmetry we look for the smallest positive angle $$\theta$$ (less than $$360^{\circ}$$) that brings the figure exactly on to itself. That $$\theta$$ is called its angle of symmetry (or angle of rotation). If no such angle smaller than $$360^{\circ}$$ exists, the figure is said to have no rotational symmetry.

Key idea 2 : Relation with a full turn
If $$\theta$$ is the smallest angle of symmetry, then repeating that rotation some whole-number $$n$$ times must give one complete turn:
\[ n\,\theta = 360^{\circ} \]
This automatically means $$\theta$$ is a factor of $$360^{\circ}$$ and the number $$n = \dfrac{360}{\theta}$$ is called the order of rotational symmetry.

(i) Statement: “Every figure will have $$360^{\circ}$$ as an angle of symmetry.”

• Any figure does coincide with itself after a full rotation of $$360^{\circ}$$, but by definition we do not count this trivial full turn when we speak of rotational symmetry.
• Figures that look the same only after $$360^{\circ}$$—for example, an irregular outline—are said to have no rotational symmetry.

Hence the statement is false.

(ii) Statement: “If the smallest angle of symmetry of a figure is a natural number in degrees, then it is a factor of $$360$$.”

Let that smallest angle be $$\theta^{\circ}$$ (with $$\theta$$ a natural number). Because turning by $$\theta^{\circ}$$ repeatedly must bring the figure back after one full turn, some whole number $$n$$ satisfies $$n\,\theta = 360$$. Therefore
\[ \theta = \dfrac{360}{n} \]
and $$\theta$$ divides $$360$$ exactly. So $$\theta$$ is indeed a factor (divisor) of 360.

The statement is true.

Pre–information about the figure
The circle printed in the book is already cut into 12 equal pizza–like slices (sectors). Name them, clockwise, as S₁, S₂, … , S₁₂. Every mirror (line of symmetry) must pass through the centre O of the circle and place each sector exactly on some other sector of the same colour.

Key idea
If a pattern has $$n$$ lines of symmetry, the whole design must repeat itself after a turn of $$\frac{360^{\circ}}{n}$$. For example, with 3 lines the design must be identical every $$120^{\circ}$$; with 4 lines, every $$90^{\circ}$$.

(i)  Three lines of symmetry

1. Group the 12 sectors into three equal blocks of 4 consecutive sectors:
      Block A : S₁, S₂, S₃, S₄
      Block B : S₅, S₆, S₇, S₈
      Block C : S₉, S₁₀, S₁₁, S₁₂
2. Inside each block make the sequence ‘red, blue, blue, red’. That is a palindrome, so the middle diameter of the block is a mirror.
3. Because the three blocks are identical and placed one after another, there are exactly 3 mirror lines (the diameters through the common edges S₂|S₃, S₆|S₇, S₁₀|S₁₁).

What to draw: colour S₁ and S₄ red, S₂ and S₃ blue; copy the same colouring for every next set of four sectors.

(ii)  Four lines of symmetry

1. Now divide the circle into four blocks of 3 consecutive sectors each.
      Block P : S₁, S₂, S₃;  Block Q : S₄, S₅, S₆;
      Block R : S₇, S₈, S₉;  Block S : S₁₀, S₁₁, S₁₂.
2. Inside every block use the palindrome ‘yellow-green-yellow’.
3. Because the same 3-sector motif is repeated four times, there is a mirror every $$90^{\circ}$$, i.e. 4 lines of symmetry (diameters through S₂, S₅, S₈, S₁₁).

What to draw: colour the 1st and 3rd sectors of each block yellow and the middle one green.

(iii)  How many different numbers of lines of symmetry are possible?

Let $$k$$ be the number of mirror lines obtained by some colouring of the 12 sectors.

• A mirror always comes in diametrically opposite pairs, so $$k$$ must be even or be 0.
• Also, when the design repeats after $$\frac{360^{\circ}}{k}$$, the repeat-angle must be a multiple of $$30^{\circ}$$ (the angle of one sector). Therefore $$k$$ must divide 12.

The divisors of 12 are 1, 2, 3, 4, 6, 12. Discard the odd 1 (a single diameter has two opposite rays, hence really 2 mirrors). Hence the attainable numbers are

0, 2, 3, 4, 6, 12.

Examples

• 0 mirrors ‑ colour every sector a different colour.
• 2 mirrors ‑ make the left half one colour and the right half another.
• 3 mirrors ‑ pattern (i) above.
• 4 mirrors ‑ pattern (ii) above.
• 6 mirrors ‑ colour alternately red, blue, red, blue, …
• 12 mirrors ‑ give all 12 sectors the same colour; the whole circle is then symmetrical about every diameter.

Hence
(i) and (ii) can be achieved by the stated colourings, and (iii) the possible counts of symmetry lines are 0, 2, 3, 4, 6 and 12.

Step 1 : Recall the two kinds of symmetry

• Reflection symmetry (also called line symmetry) means that the figure can be folded exactly on to itself along at least one straight line. Such a line is called a line of symmetry.
• Rotational symmetry means that if we turn (rotate) the figure about a fixed point through an angle that is less than $$360^{\circ}$$, the figure fits exactly on to itself. The number of different positions in which this happens in one full turn is called the order of rotational symmetry.

We must name (and be able to draw) two figures other than a circle and a square that satisfy both kinds of symmetry.

Step 2 : Pick a first figure — an equilateral triangle

1. How to draw it
   Using a ruler and a protractor (or a compass) draw a triangle whose three sides are equal. Mark its three vertices and join them.
2. Check reflection symmetry
   An equilateral triangle has three lines of symmetry. Each line passes through one vertex and the mid-point of the opposite side. Folding the triangle along any one of these lines makes the two halves coincide.
3. Check rotational symmetry
   Keep the centre of the triangle fixed and rotate it. After turning through $$120^{\circ}$$ or $$240^{\circ}$$ the triangle looks exactly the same. Because it matches its original shape three times in one full turn (at $$0^{\circ},120^{\circ},240^{\circ},360^{\circ}$$), its order of rotational symmetry is 3.

Step 3 : Pick a second figure — a regular hexagon

1. How to draw it
   With a compass draw a circle of convenient radius, keep the compass opening fixed, step the point around the circumference six times and join neighbouring points to get a hexagon whose six sides are equal.
2. Check reflection symmetry
   A regular hexagon has six lines of symmetry — three lines join opposite vertices and three lines join the mid-points of opposite sides.
3. Check rotational symmetry
   Rotating the hexagon about its centre through multiples of $$60^{\circ}$$ brings it on to itself. Hence it coincides with itself at $$0^{\circ},60^{\circ},120^{\circ},180^{\circ},240^{\circ},300^{\circ},360^{\circ}$$. Therefore, its order of rotational symmetry is 6.

Step 4 : State the required figures

An equilateral triangle and a regular hexagon, both of which are not circles or squares, possess both reflection symmetry and rotational symmetry.

Step 1 – Recall the idea of rotational symmetry
A figure is said to possess rotational symmetry if, when we rotate it about a fixed point (its centre), it fits exactly on to itself more than once in one full turn.
The angle of rotation is the amount of turning we give the figure. A complete turn measures $$360^{\circ}$$.

Step 2 – Connect the smallest angle with the order of symmetry
If the smallest angle through which the figure matches itself is $$60^{\circ}$$, then that angle must repeat a whole-number of times to make the full $$360^{\circ}$$:

\[ \text{Number of repeats} \,=\, \frac{360^{\circ}}{60^{\circ}} \,=\, 6 \]

Hence the figure has rotational symmetry of order 6. That means it will fit on to itself exactly 6 times in one full rotation.

Step 3 – List every successive multiple of the smallest angle
Starting from $$60^{\circ}$$, keep adding another $$60^{\circ}$$ until we reach the full turn $$360^{\circ}$$:

Step 4 – Identify the “other” angles
The question has already mentioned $$60^{\circ}$$, so the remaining angles of symmetry are:

• $$120^{\circ}$$
• $$180^{\circ}$$
• $$240^{\circ}$$
• $$300^{\circ}$$
• $$360^{\circ}$$ (a complete rotation)

These are exactly the angles through which the figure can be rotated and still look the same.

The question is about rotational symmetry. A figure has rotational symmetry through an angle $$\theta$$ if a turn of $$\theta$$ about its centre brings it exactly on to itself.

All such angles are always multiples of one fixed angle, called the smallest angle of symmetry. If that smallest angle is $$\theta_{\min}$$, then the whole list of symmetry angles is

$$\theta_{\min},\;2\theta_{\min},\;3\theta_{\min},\;\dots$$ up to $$360^\circ$$.

1. We are told that $$60^\circ$$ is an angle of symmetry, so it must be a multiple of the smallest one:

   $$60 = k\,\theta_{\min} \qquad (k \text{ is a positive integer}).$$

2. The figure has two symmetry angles that are smaller than $$60^\circ$$.
   From the list above, those smaller angles are $$\theta_{\min},\;2\theta_{\min},\;\dots ,\;(k-1)\theta_{\min}.$$ There are exactly $$k-1$$ of them.

   Given there are two such angles, we must have

   $$k - 1 = 2 \;\;\Longrightarrow\;\; k = 3.$$

3. Substituting $$k = 3$$ into $$60 = k\,\theta_{\min}$$ gives

   $$60 = 3\,\theta_{\min}$$

   $$\Rightarrow\; \theta_{\min} = \frac{60}{3} = 20^\circ.$$

Therefore the smallest angle of symmetry of the figure is $$20^\circ$$.

Step 1 ‒ Recall the meaning of “line of symmetry”
If a straight line divides a figure into two exactly matching halves, the line is called a line of symmetry (or axis of symmetry) of that figure.

Step 2 ‒ Observe the first shape sequence in Table 3 (Regular Polygons)
The sequence shows regular polygons with side–counts increasing one by one:

• Regular triangle (equilateral triangle)
• Square
• Regular pentagon
• Regular hexagon
• Regular heptagon
• Regular octagon

Step 3 ‒ Find the lines of symmetry one polygon at a time

Step 4 ‒ Identify the pattern
In every case the number of lines of symmetry equals the number of sides. Writing the counts in order we get:

$$3,\;4,\;5,\;6,\;7,\;8,\;\dots$$

This is simply the natural-number sequence starting from 3; it would continue as 9, 10, 11, … if we kept adding regular 9-gons, 10-gons and so on.

Step 5 ‒ General statement
For a regular polygon with $$n$$ sides, the number of lines of symmetry is also $$n$$.

Step 1 : Recall what an “angle of symmetry” means
For a plane figure an angle of rotational symmetry is an angle through which we can turn the figure about its centre so that its outline fits exactly on to itself again. One full turn is $$360^{\circ}$$, so only the fractions of this full turn that make the shape coincide count.

Step 2 : Smallest angle that works for a regular n–gon
Because all sides and all angles of a regular n-sided polygon are equal, the first time the outline matches itself is when the figure is turned through one side’s “share” of the full circle:

Step 3 : How many different such angles are there?
A second copy-fit occurs after two such turns, a third after three, and so on, until, after n such turns, we reach the full $$360^{\circ}$$ again. Hence there are exactly n rotational positions – including the starting position – in which the figure fits.

Step 4 : Apply to the regular polygons in the first row of Table 3

Step 5 : Read off the number pattern
Listing the numbers of angles of symmetry we obtain

This is exactly the sequence of natural numbers starting from 3, i.e. “three, four, five, six, …”. Each term equals the number of sides of the corresponding regular polygon.

Step 1 Recalling the construction of the Koch snow-flake

The sequence begins with an equilateral triangle. At every stage each side is divided into three equal parts and a smaller equilateral triangle is erected on the middle part. Although the boundary becomes more and more jagged, the overall shape still looks like a ‘three-armed star’ built around the centre of the original triangle.

Step 2 Looking for mirror lines (lines of symmetry)

• The first figure (the starting equilateral triangle) has 3 mirror lines: one through every vertex and the midpoint of the opposite side.
• When we add the little triangles, we do exactly the same change to each of the three sides.
  Therefore whatever happens to one side also happens, in the same way, to the other two sides that are related by the old mirror lines.
• Hence each of the original 3 mirror lines is still a mirror line of the new, more detailed outline.
• No new mirror lines are produced, because any other line would cut one ‘arm’ differently from the other two.

So the last figure of the sequence (or any stage of the Koch snow-flake) has

$$3\text{ lines of symmetry}$$

Step 3 Looking for rotational symmetry (angles of symmetry)

• Place a dot at the centre of the original triangle. Rotate the whole snow-flake about this centre.
• A rotation through $$120^{\circ}$$ moves each arm to the position of the next arm and the picture fits exactly on itself.
• Another $$120^{\circ}$$ (that is, $$240^{\circ}$$ in total) does the same thing again.
• Finally a full turn of $$360^{\circ}$$ brings it back to where it started (all figures have this). No smaller angle, apart from $$0^{\circ}$$, works.

Therefore there are

$$3\text{ angles of rotational symmetry}$$ namely $$120^{\circ},\;240^{\circ}\;\text{and}\;360^{\circ}.$$

Step 4 Conclusion

The Koch snow-flake in the last row of Table 3 keeps the same symmetry as the original equilateral triangle:

\[\boxed{\text{Lines of symmetry }=3\quad\text{ Rotational (angle) symmetries }=3}\]

The Ashoka Chakra is a circular wheel with 24 equally-spaced spokes.

(i) Lines of symmetry

• The straight line that lies on any spoke (and the exactly opposite spoke) is a mirror line. Because each axis contains a pair of opposite spokes, the number of such axes is $$\frac{24}{2}=12$$.
• A second set of mirror lines passes midway between every neighbouring pair of spokes. Again two opposite gaps share one axis, giving another $$\frac{24}{2}=12$$ axes.

Hence

\[\text{total lines of symmetry}=12+12=24.\]

(ii) Rotational (angle) symmetry

The spokes divide the full turn $$360^{\circ}$$ into 24 equal sectors:

$$\frac{360^{\circ}}{24}=15^{\circ}.$$

Every rotation through an integral multiple of $$15^{\circ}$$ brings each spoke to the position of another spoke, so the wheel looks unchanged. Thus

• smallest non-zero angle of rotational symmetry : $$15^{\circ}$$,
• order of rotational symmetry : $$24$$ (the angles are $$15^{\circ},30^{\circ},45^{\circ},\ldots,360^{\circ}$$).

Result The Ashoka Chakra has 24 lines of symmetry and rotational symmetry of order 24 with a least angle of $$15^{\circ}$$.