Exercises — Symmetry — line symmetry, rotational symmetry, order of symmetry

WHAT we do: picture a square and hunt for mirrors and turns. Lines of symmetry — 4. Look at the figure: two lines through the midpoints of opposite sides (vertical + horizontal), and two lines along the diagonals. Fold along any of these four and the halves match.

Rotational symmetry — order 4. Spinning a square about its centre, it looks identical at $0°,90°,180°,270°$. That is $4$ identical positions in one turn, so the order is $\boxed{4}$, with smallest turn $\frac{360°}{4}=90°$.

Answer: $4$ lines, order $4$.

• Line symmetry: fold left↔right — match. Fold top↔bottom — match. So $2$ lines of symmetry.
• Rotational symmetry: turn it $180°$ about its middle point and the H is unchanged. That is $2$ identical positions ($0°$ and $180°$), so order $\boxed{2}$.

Answer: $2$ lines of symmetry; rotational order $2$.

Only the full turn works, so there is exactly one identical position (the start). Order $=\boxed{1}$. By convention, order $1$ means the figure has no rotational symmetry — the $360°$ turn always works for everything, so it counts for nothing special.

WHY this tool: the mirror is vertical ($x=7$), so heights ($y$) never change and only the horizontal position flips. That is exactly what $P^{\prime }(2a-x,y)$ says, with $a=7$.

• $A(2,1)$: $x^{\prime }=2(7)-2=12$, so $A^{\prime }(12,1)$. (A is $5$ left of the mirror, $A^{\prime }$ is $5$ right.)
• $B(5,3)$: $x^{\prime }=14-5=9$, so $B^{\prime }(9,3)$. ($2$ left → $2$ right.)
• $C(3,6)$: $x^{\prime }=14-3=11$, so $C^{\prime }(11,6)$. ($4$ left → $4$ right.)

Check distances: $|2-7|=|12-7|=5$ ✓, $|5-7|=|9-7|=2$ ✓, $|3-7|=|11-7|=4$ ✓

Answer: $A^{\prime }(12,1),B^{\prime }(9,3),C^{\prime }(11,6)$.

A regular $n$-gon has $n$ lines of symmetry and rotational order $n$. Here $n=8$.

• Lines of symmetry: $8$ (through $4$ pairs of opposite vertices + through $4$ pairs of opposite side-midpoints).
• Rotational order: $8$ (centre = middle of the polygon).
• Smallest turn: $\frac{360°}{8}=45°$.

Answer: $8$ lines, order $8$, turn $45°$.

WHY a different formula: the mirror is horizontal now ($y=1$), so the left–right position ($x$) stays fixed and only the height flips. That is $P^{\prime }(x,2b-y)$ with $b=1$. $x^{\prime }=-3,y^{\prime }=2b-y=2(1)-4=2-4=-2$ So $P^{\prime }(-3,-2)$. Check: $P$ is at $y=4$, the mirror at $y=1$, distance $=|4-1|=3$. $P^{\prime }$ is at $y=-2$, distance $=|-2-1|=3$ ✓ — same distance, opposite side of the horizontal mirror.

Answer: $P^{\prime }(-3,-2)$.

Test each separately — folding for lines, spinning for order.

Why X has order 4: an ideal X is two equal strokes crossing at right angles, so it is really the same shape as a "plus" sign tilted $45°$. Fold it vertically, horizontally, or along either diagonal — every fold matches, giving $4$ axes. Spinning it about the crossing point, it looks identical at $90°,180°,270°$, so rotational order $4$. Key contrast: N and S have rotational symmetry but no line symmetry — proof that the two symmetries are independent — while X, like a regular polygon, has lines equal to order.

WHY divide: the order counts how many equal slices of $40°$ fit in a full turn. $\text{order}=\frac{360°}{40°}=9$ The matching angles are the multiples of $40°$ up to the full turn: $40°,80°,120°,160°,200°,240°,280°,320°,360°$ That is $9$ angles, matching order $9$. ✓

Answer: order $9$; angles $40°,80°,\dots ,360°$.

A regular hexagon has order $6$, smallest turn $\frac{360°}{6}=60°$. It maps onto itself at every multiple of $60°$: $60°,120°,180°,240°,300°,360°$. Since $120°=2\times 60°$ is such a multiple, yes, a $120°$ turn maps the hexagon onto itself. General rule: an order-$n$ figure is unchanged by a turn of $\theta$ exactly when $\theta$ is a whole-number multiple of $\frac{360°}{n}$.

Answer: Yes — $120°$ is $2\times 60°$, a multiple of the smallest turn.

Equilateral triangle: $3$ lines of symmetry (each from a vertex to the midpoint of the opposite side — these are also its medians/altitudes/angle bisectors, since all coincide here). Rotational order $3$ about its centroid, smallest turn $\frac{360°}{3}=120°$.

Verify the claim (lines vs order):

Answer: Triangle has $3$ lines, order $3$; the claim lines $=$ order holds for all three.

Construction — a "pinwheel". Take a square and attach to each side an identical bent flag, all bending the same way (say clockwise). Because all four arms are identical and equally spaced, a $90°$ turn about the centre carries each arm onto the next — order $4$. But every candidate mirror line (vertical, horizontal, or diagonal) would have to flip the bend direction (clockwise becomes anticlockwise), so no fold matches the original. Hence $0$ lines of symmetry. This is the deep lesson: rotation preserves "handedness", reflection reverses it. A shape with a built-in swirl can spin onto itself but never fold onto itself.

Answer: A same-handed pinwheel — order $4$, $0$ lines.

(a) Smallest turn $=\frac{360°}{12}=30°$.

(b) It maps onto itself only at multiples of $30°$. Is $150°$ a multiple of $30°$? $150/30=5$, a whole number — yes, it maps onto itself ($150°=5\times 30°$).

(c) The matching angles are $30°,60°,90°,120°,\dots$. Discard the multiples of $90°$ ($90°,180°,\dots$). The smallest matching angle that is not a multiple of $90°$ is the very first one, $\boxed{30°}$ (since $30°$ is not divisible by $90°$).

Answer: (a) $30°$; (b) yes; (c) $30°$.

First reflection ($a=1$): $x^{\prime }=2(1)-4=-2$, so $P^{\prime }(-2,7)$. Second reflection ($a=5$): $x^{\prime \prime }=2(5)-(-2)=10+2=12$, so $P^{\prime \prime }(12,7)$.

The slide: original $x=4$ became $x=12$, a change of $12-4=+8$; $y$ is unchanged. So the two reflections combine into a single horizontal shift of $8$ units to the right (a translation). WHY $8$: two parallel mirrors at distance $d=5-1=4$ apart always produce a translation of $2d=8$, in the direction from the first mirror to the second. Here the second mirror ($x=5$) is to the right of the first ($x=1$), so the slide is rightward. This is a cornerstone fact of reflections composing into translations.

Answer: $P^{\prime \prime }(12,7)$; equivalent to a translation of $8$ units right.

(a) Smallest turn $=\frac{360°}{n}=18°\Rightarrow n=\frac{360}{18}=20$. So a regular $20$-gon. (b) For a regular $n$-gon, lines of symmetry $=n=20$.

Answer: (a) $20$ sides; (b) $20$ lines of symmetry.

Reflection flips handedness (fold test; formula depends on the mirror's direction — $2a-x$ for vertical $x=a$, $2b-y$ for horizontal $y=b$, swap for diagonal $y=x$); rotation preserves it (spin test about the centroid, order $=360°/\text{turn}$). They are independent, and an angle matches an order-$n$ figure exactly when it is a whole multiple of $360°/n$.