Symmetry Worked ExamplesIGCSE Maths 0580

(a) A square has two diagonals and two lines through opposite edge midpoints, and maps onto itself every $90^{\circ}$.

(b) A rectangle has only the two lines through opposite edge midpoints — its diagonals are not lines of symmetry.

(c) An equilateral triangle has one line per vertex.

(d) A parallelogram has no line of symmetry at all, but a half-turn maps it onto itself.

(a) A regular hexagon has three lines through opposite vertices and three through opposite edge midpoints.

(b) Every regular polygon matches its side count in both measures.

(a) Each plane cuts the cuboid halfway along one of the three directions.

(b) A cube keeps those three, and because all its faces are identical squares it gains six more — one through each pair of opposite edges, cutting diagonally.

Two lines of symmetry and order $2$ narrows it to a rectangle or a rhombus — a square would have four lines.

A rectangle has equal diagonals, so the unequal-diagonals condition rules it out.

A rhombus has two lines of symmetry along its diagonals, order $2$, and unequal diagonals.

(a) Divide a full turn by the order.

(b) Every multiple of that angle also works, up to the full turn.

The three points form a right angle at $(4, 1)$, with a horizontal side of length $3$ and a vertical side of length $2$.

(a) To close it into a rectangle, the fourth vertex must sit above $(1,1)$ at the same height as $(4,3)$.

(b) The sides are $3$ and $2$ — unequal, so it is a rectangle rather than a square.

A rectangle has two lines of symmetry and order $2$, as required.

(a) Order $2$ means the pattern looks the same after a half turn ($180^{\circ}$) about $O$. Rotate each shaded square through $180^{\circ}$: it moves to the square diagonally opposite it through $O$.

Shade $D1$, $C1$ and $C2$. Check that a quarter turn does not map the new pattern onto itself (for example, $A4$ would move to $D4$, which is not shaded), so the order is exactly $2$.

(b) Test each possible mirror line of the square grid: the vertical and horizontal lines through $O$ and the two diagonals. The vertical line sends $A4$ to $D4$ (not shaded), the horizontal line sends $A4$ to $A1$ (not shaded), the diagonal from $A1$ to $D4$ sends $B4$ to $D2$ (not shaded), and the diagonal from $A4$ to $D1$ sends $B4$ to $A3$ (not shaded).

(c) For order $4$, every quarter turn must map the pattern onto itself, so each shaded square needs all four of its positions shaded. Follow each square round in quarter turns.

That is $3 \times 4 = 12$ squares altogether, and $3$ are already shaded. None of the positions repeat, so no smaller number works.

(a) $OA = OB$ (radii), so the line from $O$ perpendicular to $AB$ is the line of symmetry of triangle $OAB$ and of the chord. The perpendicular bisector of a chord passes through the centre, so the perpendicular from $O$ meets $AB$ at its midpoint.

(b) Triangle $OMA$ has a right angle at $M$ and hypotenuse $OA = 13\text{ cm}$.

(c) In the same way, $N$ is the midpoint of $CD$, so $CN = 5\text{ cm}$. Draw the radius $OC$ ($13\text{ cm}$) and use triangle $ONC$.

The chords are on opposite sides of $O$, so the distance between them is the sum.

(d) $EF$ is the same length as $AB$. Equal chords are equidistant from the centre, so $EF$ is the same distance from $O$ as $AB$.

(a)(i) A regular hexagon has $6$ lines of symmetry: $3$ through opposite vertices and $3$ through the midpoints of opposite sides. Each one, extended up through the prism, gives a vertical plane of symmetry. There is also one horizontal plane halfway up, which swaps the top and bottom faces.

(a)(ii) Turning about the axis matches the hexagon to itself every $60^{\circ}$.

(b) Each of the $5$ lines of symmetry of the regular pentagon gives a vertical plane through the vertex of the pyramid. There is no horizontal plane — it would have to swap the pointed top with the flat base. About the vertical axis the pyramid matches itself every $72^{\circ}$.

(c) A regular $n$-sided polygon has $n$ lines of symmetry, and each gives a vertical plane of symmetry of the prism. The plane halfway along the prism, parallel to the two end faces, is one more. That makes $n + 1$.

(d) A cube is a prism with a square cross-section. Its length equals the side of the square, so extra planes appear: $3$ planes parallel to pairs of faces and $6$ planes through pairs of opposite edges.

(a) A vertical or horizontal mirror line cannot work with only one extra square: $B1$ and $B2$ would need partners in column $C$, and $D3$ a partner in column $A$ or row $2$ — that is at least two more squares. So try the diagonals of the grid.

Take the diagonal from the top-left corner of $A4$ to the bottom-right corner of $D1$. Reflecting in it swaps columns and rows: $B1 \leftrightarrow D3$, which are both shaded already, and $B2 \leftrightarrow C3$.

Check that there is only one line: the other diagonal sends $B1$ to $A2$, which is not shaded, so the pattern $B1, B2, C3, D3$ has exactly one line of symmetry.

(b) With $6$ columns and $5$ rows, the only possible mirror lines are the vertical line between columns $C$ and $D$ and the horizontal line through row $3$. For both, every shaded square in column $B$ needs a partner in column $E$ (vertical line), and $E5$ needs a partner in row $1$ (horizontal line).

Shading $E1$ and $E3$ gives two columns of three squares, which is symmetrical in both lines.

(a) Every plane of symmetry of the pyramid must pass through $T$ and cut the base along one of its lines of symmetry. A rectangle that is not a square has only $2$ lines of symmetry (its diagonals are not lines of symmetry).

(b) The axis is the vertical line about which the pyramid can be turned onto itself. It passes through the vertex and the centre of the base.

(c) A half turn about $TN$ maps the rectangle onto itself; a quarter turn does not, because the sides are different lengths.

(d) A prism whose cross-section is a regular $n$-sided polygon has $n$ planes along its length, one for each line of symmetry of the polygon, plus $1$ plane across the middle.

A regular polygon with $10$ sides is a regular decagon.