Symmetry

A rectangle has two lines of symmetry, each of which is the line joining the mid-points of a pair of its opposite sides.

A semicircle has only one line of symmetry, which is the perpendicular bisector of its diameter.

A regular pentagon has five lines of symmetry, each of which is the perpendicular bisector of one of its sides.

A rhombus has two lines of symmetry, which are its two diagonals.

Hence, only a semicircle has exactly one line of symmetry.

A square has four lines of symmetry, which are its two diagonals and the two lines joining the mid-points of its opposite sides.

A parallelogram has no line of symmetry.

An isosceles trapezium has only one line of symmetry, which is the line joining the mid-points of its two parallel sides.

A rectangle has two lines of symmetry, each of which is the line joining the mid-points of a pair of its opposite sides.

Hence, only a rectangle has exactly two lines of symmetry.

In an isosceles triangle only two sides are equal. It can be folded into two coincident parts only about the bisector of the angle included between its equal sides, which is also the perpendicular bisector of its third side.

Hence, an isosceles triangle has one line of symmetry.

A square has four lines of symmetry, which are its two diagonals and the two lines joining the mid-points of its opposite sides.

A rectangle has two lines of symmetry, each of which is the line joining the mid-points of a pair of its opposite sides. On folding a rectangle about either of its diagonals, the two parts do not coincide, so its diagonals are not its lines of symmetry.

Hence, a square and a rectangle have an unequal number of lines of symmetry.

The letter C has one line of symmetry, which is horizontal.

The letter M has one line of symmetry, which is vertical.

The letter B has one line of symmetry, which is horizontal.

The letter S cannot be divided into two coincident parts by any line, so it does not possess a linear symmetry. It possesses only a point symmetry.

Hence, S does not possess a linear symmetry.

(i) The given figure is made up of a pointed dome resting on a stem, which stands on a rectangular base. It can be folded into two coincident parts only about the vertical line drawn through the tip of the dome and the mid-point of the base.

Hence, the given figure has 1 line of symmetry.

(ii) The given figure is a circle divided into two parts by an S-shaped curve. There is no line about which the figure can be folded so that its two parts coincide.

Hence, the given figure has no line of symmetry.

(iii) The given figure is a quadrant of a circle. It can be folded into two coincident parts only about the bisector of the right angle at its centre.

Hence, the given figure has 1 line of symmetry.

(iv) The given figure is a five-pointed star made up of five identical points. It can be folded into two coincident parts about each of the five lines drawn through a point of the star and the inner corner opposite to it.

Hence, the given figure has 5 lines of symmetry.

Steps of construction:

1. Draw QR = 4 cm.

2. At Q, construct ∠RQX = 45°.

3. At R, construct ∠QRY = 90°.

4. Let the rays QX and RY intersect at the point P.

In △PQR, we have :

∠P + ∠Q + ∠R = 180°

⇒ ∠P + 45° + 90° = 180°

⇒ ∠P = 180° − 135° = 45°

∴ ∠P = ∠Q = 45°, so PR = QR, i.e., PQR is an isosceles triangle.

It can be folded into two coincident parts only about the bisector of ∠R, which is also the perpendicular bisector of PQ. It is shown by the dotted line in the figure.

Hence, PQR is the required triangle and it has 1 line of symmetry.

Steps of construction:

1. Draw XY = 3.5 cm.

2. At X, construct ∠YXP = 65°.

3. At Y, construct ∠XYQ = 65°.

4. Let the rays XP and YQ intersect at the point Z.

In △XYZ, we have :

∠X + ∠Y + ∠Z = 180°

⇒ 65° + 65° + ∠Z = 180°

⇒ ∠Z = 180° − 130° = 50°

∴ ∠X = ∠Y = 65°, so ZY = ZX, i.e., XYZ is an isosceles triangle.

It can be folded into two coincident parts only about the bisector of ∠Z, which is also the perpendicular bisector of XY. It is shown by the dotted line in the figure.

Hence, XYZ is the required triangle and it has 1 line of symmetry.

Steps of construction:

1. Draw QR = 6.5 cm.

2. At Q, construct ∠XQR = 80°.

3. From QX, cut off QP = 6.5 cm.

In ∠PQR, we have :

QP = QR = 6.5 cm

Let QS be the bisector of ∠PQR.

⇒ ∠PQS = ∠SQR = $\frac{80^{\circ}}{2}$ = 40°

∴ QP = QR and ∠PQS = ∠SQR = 40°, so on folding the figure about QS, the arm QP falls exactly over the arm QR.

It can be folded into two coincident parts only about the bisector QS of ∠PQR. It is shown by the dotted line in the figure.

Hence, ∠PQR = 80° is the required angle and it has 1 line of symmetry.

We know that,

A figure possesses a point symmetry about a point O if it attains its original form on being rotated through 180° about O, and if A° (≤ 180°) is the smallest angle through which a figure has to be rotated about a fixed point so that it looks exactly the same as in its starting position, then the order of its rotational symmetry = $\frac{360^{\circ}}{A^{\circ}}$.

(i) The given figure is made up of three identical blades meeting at the point O.

Reflection Symmetry: There is no line about which the figure can be folded so that its two parts coincide. So, the figure has no line of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it does not look the same as in its starting position. So, the figure has no point of symmetry.

Rotational Symmetry: On rotating the figure through 120° about the point O, it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{120^{\circ}} = 3$.

Hence, the given figure has no line of symmetry, no point of symmetry and a rotational symmetry of order 3 about the point O.

(ii) The given figure is made up of a rectangle AEFJ having an identical triangular projection BCD on its upper side and IHG on its lower side.

Reflection Symmetry: The figure can be folded into two coincident parts about the line joining the points C and H, and also about the line joining the mid-points of AJ and EF. So, the figure has two lines of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it looks exactly the same as in its starting position. So, the point O is the centre of symmetry of the figure.

Rotational Symmetry: The smallest angle through which the figure has to be rotated about the point O to look exactly the same is 180°.

Order of rotational symmetry = $\frac{360^{\circ}}{180^{\circ}} = 2$.

Hence, the given figure has two lines of symmetry — the line joining C and H and the line joining the mid-points of AJ and EF, a point symmetry about the point O and a rotational symmetry of order 2 about the point O.

(iii) The given figure is a square ABCD divided into two identical step-shaped parts by a stepped line drawn through the point O.

Reflection Symmetry: There is no line about which the figure can be folded so that its two parts coincide. So, the figure has no line of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it looks exactly the same as in its starting position. So, the point O is the centre of symmetry of the figure.

Rotational Symmetry: The smallest angle through which the figure has to be rotated about the point O to look exactly the same is 180°.

Order of rotational symmetry = $\frac{360^{\circ}}{180^{\circ}} = 2$.

Hence, the given figure has no line of symmetry, a point symmetry about the point O and a rotational symmetry of order 2 about the point O.

(iv) The given figure is made up of three identical petals meeting at the point O, with A, C and E as the tips of the petals and B, D and F as the corners between them.

Reflection Symmetry: The figure can be folded into two coincident parts about the line joining A and D, about the line joining B and E and about the line joining C and F. So, the figure has three lines of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it does not look the same as in its starting position. So, the figure has no point of symmetry.

Rotational Symmetry: On rotating the figure through 120° about the point O, it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{120^{\circ}} = 3$.

Hence, the given figure has three lines of symmetry — the lines joining AD, BE and CF, no point of symmetry and a rotational symmetry of order 3 about the point O.

(v) The given figure is a triangle ABC drawn in the form of a spiral made up of three identical bands.

Reflection Symmetry: There is no line about which the figure can be folded so that its two parts coincide. So, the figure has no line of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it does not look the same as in its starting position. So, the figure has no point of symmetry.

Rotational Symmetry: On rotating the figure through 120° about the point O, it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{120^{\circ}} = 3$.

Hence, the given figure has no line of symmetry, no point of symmetry and a rotational symmetry of order 3 about the point O.

(vi) The given figure is made up of two equal parallel line segments AB and CD joined by the line segment BC, with O as the mid-point of BC.

Reflection Symmetry: There is no line about which the figure can be folded so that its two parts coincide. So, the figure has no line of symmetry.

Point Symmetry: On rotating the figure through 180° about the point O, it looks exactly the same as in its starting position. So, the point O is the centre of symmetry of the figure.

Rotational Symmetry: The smallest angle through which the figure has to be rotated about the point O to look exactly the same is 180°.

Order of rotational symmetry = $\frac{360^{\circ}}{180^{\circ}} = 2$.

Hence, the given figure has no line of symmetry, a point symmetry about the point O and a rotational symmetry of order 2 about the point O.

The given figure is a square-wave shaped pattern drawn between two shaded strips. On rotating it through 180° about its centre, it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{180^{\circ}} = 2$.

So, Assertion (A) is true.

Every figure attains its original form on being rotated through 360° about a fixed point, that is, every figure fits onto itself at least once in the process of a complete rotation. So, each figure has a rotational symmetry of order 1.

So, Reason (R) is true.

But the order of rotational symmetry of the given figure is 2 because it fits onto itself on being rotated through 180°, and not because every figure has a rotational symmetry of order 1. So, Reason (R) is not the correct explanation of Assertion (A).

∴ Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

A regular hexagon has a rotational symmetry of order 6 about its centre.

∴ Minimum angle of rotation = $\frac{360^{\circ}}{6} = 60^{\circ}$

So, on being rotated through 60° about its centre, a regular hexagon coincides with its original position for the first time.

So, Assertion (A) is true.

A figure that possesses a point symmetry attains its original form upon being rotated through 180° about its centre of symmetry, and not through 120°.

So, Reason (R) is false.

∴ Assertion (A) is true but Reason (R) is false.

Hence, option 3 is the correct option.

The given figure is a square having four identical circles drawn near its four corners and one circle drawn at its centre.

On rotating the figure through 90° about its centre, the circle at the centre remains in its own place and each of the four corner circles takes the place of the next corner circle, so the figure looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{90^{\circ}} = 4$.

Hence, option 2 is the correct option.

The given figure is made up of six identical bent arms placed around its centre in the shape of a hexagon.

On rotating the figure through 60° about its centre, each arm takes the place of the next arm, so the figure looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{60^{\circ}} = 6$.

Hence, option 2 is the correct option.

The given figure is a border pattern made up of two identical bent designs drawn between two straight strips.

There is no line about which the figure can be folded so that its two parts coincide. So, the figure has no line of symmetry.

On rotating the figure through 180° about its centre, it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{180^{\circ}} = 2$.

So, the given figure has no line of symmetry and a rotational symmetry of order 2.

Hence, option 3 is the correct option.

We know that,

If A° is the smallest angle through which a figure has to be rotated so that it looks exactly the same as in its starting position, then its order of rotational symmetry = $\frac{360^{\circ}}{A^{\circ}}$, which must be a whole number.

For 32°, order of rotational symmetry = $\frac{360^{\circ}}{32^{\circ}}$ = 11.25, which is not a whole number.

So, there can be no figure whose smallest angle of rotational symmetry is 32°, and the student's statement is incorrect.

Hence, option 2 is the correct option.

The Ashok Chakra is a wheel having 24 identical spokes placed at equal distances around its centre.

∴ Angle between two consecutive spokes = $\frac{360^{\circ}}{24}$ = 15°

On rotating the Ashok Chakra through 15° about its centre, each spoke takes the place of the next spoke, so it looks exactly the same as in its starting position.

Order of rotational symmetry = $\frac{360^{\circ}}{15^{\circ}} = 24$.

Hence, option 4 is the correct option.

For the completed grid to have a rotational symmetry of order 4, it must look exactly the same on being rotated through 90° about the centre of the grid.

On rotating the grid through 90° in the clockwise direction about its centre :

the square 3 takes the place of the square 12,

the square 12 takes the place of the square 14,

the square 14 takes the place of the square 5,

and the square 5 takes the place of the square 3.

So, along with the square 3, the squares 5, 12 and 14 must also be shaded.

Hence, option 3 is the correct option.

In the given grid, the squares 3 and 10 are shaded.

Consider the diagonal of the grid drawn through its top right corner and its bottom left corner, i.e., the diagonal passing through the squares 4, 7, 10 and 13.

The square 10 lies on this diagonal, so it falls on itself when the grid is folded about it. The square 3 falls exactly on the square 8.

So, on shading the square 8, the grid can be folded into two coincident parts about this diagonal.

Hence, option 1 is the correct option.