Question

1. draw all lines of reflection for the 6 - pointed star shown here: use straight edge
2. what is the smallest angle of rotation so that the image lays perfectly over the original star?
3. describe a sequence of rigid transformations that will take abcd onto a’b’c’d’.

note: there is more than one right way, but your list of steps has to work! give all info. (don’t just say “rotate” for example. rotate how?) we may ask you to show us your steps using tracing paper.
list your steps here, as many as you need:

1. find the measure of each angle (less than 180 deg.). show, or tell, how you know.

why?
m∠abc = ____
m∠abe = ____
m∠ebd = ____

Explanation:

Step1: Lines of reflection for 6 - pointed star

A 6 - pointed star has 6 lines of reflection. These lines pass through the opposite vertices and the mid - points of opposite sides of the star.

Step2: Smallest rotation angle for 6 - pointed star

The formula for the smallest angle of rotation $\theta$ for a regular $n$ - sided figure is $\theta=\frac{360^{\circ}}{n}$. For a 6 - pointed star (which has 6 fold rotational symmetry), $n = 6$. So $\theta=\frac{360^{\circ}}{6}=60^{\circ}$.

Step3: Rigid transformations for rectangle

1. First, translate rectangle $ABCD$ so that point $A$ coincides with point $A'$.
2. Then, rotate the translated rectangle about point $A'$ by the angle between $AB$ and $A'B'$ until $AB$ aligns with $A'B'$.

Step4: Angle measures

• $\angle ABC$ and the $40^{\circ}$ angle are supplementary (linear pair). So $m\angle ABC = 180^{\circ}-40^{\circ}=140^{\circ}$.
• $\angle ABE$ and $\angle ABC$ are supplementary. So $m\angle ABE=40^{\circ}$ (since $\angle ABE + \angle ABC=180^{\circ}$).
• $\angle EBD$ and the $40^{\circ}$ angle are vertical angles. So $m\angle EBD = 40^{\circ}$ (vertical angles are equal).

Answer:

1. 6 lines of reflection (passing through opposite vertices and mid - points of opposite sides).
2. $60^{\circ}$
3. 1. Translate so that $A$ coincides with $A'$. 2. Rotate about $A'$ to align $AB$ with $A'B'$.
4. $m\angle ABC = 140^{\circ}$, $m\angle ABE = 40^{\circ}$, $m\angle EBD = 40^{\circ}$