Symmetry, Reflection and Rotation

(i) The given shape is made up of two circles of unequal radii. The figure can be folded into two coincident parts only about the line passing through the centres of both the circles.

Hence, the given shape has 1 line of symmetry.

(ii) The given shape is made up of a smaller circle lying inside a bigger circle and touching it at one point. The figure can be folded into two coincident parts only about the line passing through the centres of both the circles.

Hence, the given shape has 1 line of symmetry.

(iii) The given shape 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 shape has no line of symmetry.

(iv) The given shape is a cross. It can be folded into two coincident parts about each of the two lines joining the mid-points of its opposite arms and about each of the two lines drawn through the centre making an angle of 45° with these arms.

Hence, the given shape has 4 lines of symmetry.

(v) The given shape is made up of three identical petals meeting at a point. It can be folded into two coincident parts about the line drawn along the axis of each petal.

Hence, the given shape has 3 lines of symmetry.

(vi) The given shape is made up of four identical petals meeting at a point. It can be folded into two coincident parts about each of the two lines drawn along the axes of opposite petals and about each of the two lines drawn through the meeting point, bisecting the angles between adjacent petals.

Hence, the given shape has 4 lines of symmetry.

We know that,

If A° (≤ 180°) is the smallest angle through which a figure can be rotated about a fixed point and still looks the same as in its starting position, then the order of its rotational symmetry = $\frac{360^{\circ}}{A^{\circ}}$.

(i) The two circles forming the shape are of unequal radii. So, the shape comes back to its starting position only after a complete rotation of 360°.

Hence, the given shape has no rotational symmetry.

(ii) The two circles forming the shape are of unequal radii and they touch each other at one point. So, the shape comes back to its starting position only after a complete rotation of 360°.

Hence, the given shape has no rotational symmetry.

(iii) On rotating the given shape through 180° about the centre of the circle, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 2.

(iv) On rotating the given shape through 90° about the point of intersection of its arms, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 4.

(v) On rotating the given shape through 120° about the point where its three petals meet, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 3.

(vi) On rotating the given shape through 90° about the point where its four petals meet, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 4.

Steps of construction:

1. Draw AB = 4.5 cm.

2. At B, construct ∠XBA = 90°. From BX cut off BC = 3 cm.

3. Taking C as centre and radius 4.5 cm, and A as centre and radius 3 cm, draw arcs on the required side to intersect at D.

4. Join AD and CD.

A rectangle has two lines of symmetry — the lines joining the mid-points of its opposite sides. They are shown by the dotted lines in the figure above.

Hence, ABCD is the required rectangle and it has 2 lines of symmetry.

We know that,

All the sides of a rhombus are equal.

∴ AB = BC = CD = DA = 5.3 cm

Steps of construction:

1. Draw AB = 5.3 cm.

2. At A, construct ∠XAB = 60°. From AX cut off AD = 5.3 cm.

3. Taking B and D as centres and radius 5.3 cm, draw arcs to intersect again at C (other than A).

4. Join BC and DC.

A rhombus has two lines of symmetry — its diagonals AC and BD. They are shown by the dotted lines in the figure above.

Hence, ABCD is the required rhombus and it has 2 lines of symmetry.

(i) A figure has line symmetry if it is its own image under a reflection.

(ii) A kite has one line(s) of symmetry.

(iii) A parallelogram has no line(s) of symmetry.

(iv) The centre of rotation of an equilateral triangle is the point of intersection of its angle bisectors/medians/altitudes.

(v) The centre of rotation of a rhombus is the point of intersection of its diagonals.

(vi) A regular polygon of $n$-sides has $\bm{n}$ number of lines of symmetry.

(vii) Angle of rotational symmetry in an equilateral triangle is 120°.

(viii) Angle of rotational symmetry in a regular pentagon is 72°.

(ix) If after a rotation of 45° about a fixed point the figure looks exactly the same, then the order of rotational symmetry is 8.

(i) False

Reason:

A parallelogram has no line of symmetry. Each diagonal divides it into two congruent triangles, but on folding the parallelogram about a diagonal the two parts do not coincide.

(ii) False

Reason:

A regular triangle is an equilateral triangle. It has three lines of symmetry, which are the bisectors of its angles, and rotational symmetry of order 3, but it has no point of symmetry.

(iii) True

Reason:

A regular quadrilateral is a square. It has four lines of symmetry, which are the lines joining the mid-points of its opposite sides and its diagonals, one point of symmetry, which is the point of intersection of its diagonals, and rotational symmetry of order 4.

(iv) False

Reason:

A parallelogram looks exactly the same on being rotated through 180° about the point of intersection of its diagonals. So, it has rotational symmetry of order 2.

(v) False

Reason:

A regular pentagon has no point of symmetry, as on rotating it through 180° about its centre it does not look the same as in its starting position.

(vi) False

Reason:

The letter Z has no line of symmetry. It has one point of symmetry and rotational symmetry of order 2.

A kite and an isosceles trapezium are quadrilaterals having 1 line of symmetry.

A rectangle and a rhombus are quadrilaterals having 2 lines of symmetry.

A square is a quadrilateral having 4 lines of symmetry.

There is no quadrilateral having exactly 3 lines of symmetry.

Hence, option 3 is the correct option.

We know that,

If A° is the smallest angle through which a figure can be rotated and still looks the same, then its order of rotational symmetry = $\frac{360^{\circ}}{A^{\circ}}$, which must be a whole number greater than or equal to 2.

For 36°, order = $\frac{360^{\circ}}{36^{\circ}} = 10$, which is a whole number greater than 2.

For 144°, order = $\frac{360^{\circ}}{144^{\circ}} = 2.5$, which is not a whole number.

For 150°, order = $\frac{360^{\circ}}{150^{\circ}} = 2.4$, which is not a whole number.

For 360°, order = $\frac{360^{\circ}}{360^{\circ}} = 1$, which means the figure has no rotational symmetry.

So, the possible angle of rotation is 36°.

Hence, option 1 is the correct option.

The figure in option 1 is a cross. It has 4 lines of symmetry and rotational symmetry of order 4.

The figure in option 2 is a circle with two perpendicular diameters. It has 4 lines of symmetry and rotational symmetry of order 4.

The figure in option 4 is made up of three identical petals. It has 3 lines of symmetry and rotational symmetry of order 3.

The figure in option 3 is made up of three identical bent arms meeting at a point. On rotating it through 120° about this point it looks exactly the same, so it has rotational symmetry of order 3. But there is no line about which it can be folded into two coincident parts, so it has no line of symmetry.

Hence, option 3 is the correct option.

The letter H has two lines of symmetry, one vertical and one horizontal, and it also looks exactly the same on being rotated through 180° about its centre. So, it has rotational symmetry of order 2.

The letter M has one vertical line of symmetry but no rotational symmetry.

The letter S has rotational symmetry of order 2 but no line of symmetry.

The letter Y has one vertical line of symmetry but no rotational symmetry.

So, the letter H has both line and rotational symmetry.

Hence, option 1 is the correct option.

A rhombus has two lines of symmetry, which are its diagonals. On folding the rhombus about either of its diagonals, the two parts coincide. So, the diagonals of a rhombus are its mirror lines.

So, Statement I is true.

On rotating a rhombus through 180° about the point of intersection of its diagonals, it looks exactly the same as in its starting position.

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

So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

A figure has a point of symmetry if it looks exactly the same on being rotated through 180° about that point.

On rotating a regular pentagon through 180° about its centre, it does not look the same as in its starting position. So, a regular pentagon has no point of symmetry.

So, Statement I is false.

On rotating a regular hexagon through 180° about its centre, it looks exactly the same as in its starting position. So, a regular hexagon has one point of symmetry, which is its centre.

So, Statement II is true.

∴ Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

The figure given is the letter S. On folding the letter S about the dotted line, the two parts do not coincide. In fact, the letter S has no line of symmetry but it has a point of symmetry.

So, Statement I is false.

A plane figure is symmetrical about a line if that line divides it into two coincident parts, and such a line is called its line of symmetry. This is the definition of line symmetry.

So, Statement II is true.

∴ Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

A regular hexagon has rotational symmetry of order 6.

∴ Angle of rotational symmetry = $\frac{360^{\circ}}{6} = 60^{\circ}$, and not 30°.

So, Statement I is false.

If A° (≤ 180°) is the smallest angle through which a figure can be rotated and still looks the same, then it has a rotational symmetry of order $\frac{360^{\circ}}{A^{\circ}}$ and not $\frac{180^{\circ}}{A^{\circ}}$.

So, Statement II is false.

∴ Both Statement I and Statement II are false.

Hence, option 4 is the correct option.

(i) The given shape is an ellipse. It can be folded into two coincident parts about its longer axis and about its shorter axis.

Hence, the given shape has 2 lines of symmetry.

(ii) The given shape is made up of two identical circles placed above and below a rectangular bar. It can be folded into two coincident parts about the vertical line through the centres of the two circles and about the horizontal line through the centre of the bar.

Hence, the given shape has 2 lines of symmetry.

(iii) The given shape is made up of three identical bars joined to one another. There is no line about which the figure can be folded so that its two parts coincide.

Hence, the given shape has no line of symmetry.

We know that,

If A° (≤ 180°) is the smallest angle through which a figure can be rotated about a fixed point and still looks the same as in its starting position, then the order of its rotational symmetry = $\frac{360^{\circ}}{A^{\circ}}$.

(i) On rotating the given shape through 180° about the point of intersection of its two axes, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 2.

(ii) On rotating the given shape through 180° about the centre of the bar, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 2.

(iii) On rotating the given shape through 120° about its centre, it looks exactly the same as in its starting position.

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

Hence, the given shape has rotational symmetry of order 3.

The word 'NEW ALGEBRA' is made up of the letters N, E, W, A, L, G, E, B, R and A.

The letters E and B each have one horizontal line of symmetry.

The letters W and A each have one vertical line of symmetry.

The letters N, L, G and R cannot be divided into two coincident parts by any line, so they have no line of symmetry.

Hence, the letters N, L, G and R of the word 'NEW ALGEBRA' have no line of symmetry.