Question

select the correct answer
regular hexagon fghijk shares a common center with square abcd on a coordinate plane. (overline{ab}paralleloverline{fg}). across which lines can the combined figure
reflect onto itself?
a. any of the perpendicular bisectors of the sides of the hexagon
b. either diagonal of the square
c. either perpendicular bisector of the sides of the square
d. there are no lines across which this figure can reflect onto itself

Explanation:

Step1: Analyze the symmetry of the combined figure

A regular hexagon has 6 lines of symmetry (the perpendicular bisectors of its sides). A square has 4 lines of symmetry: 2 perpendicular bisectors of its sides and 2 diagonals. Since \(\overline{AB}\parallel\overline{FG}\), the combined figure's symmetry is determined by the common - symmetric lines.
The perpendicular bisectors of the sides of the square are also lines of symmetry for the combined figure. When we reflect the combined figure across the perpendicular bisector of a side of the square, the square maps onto itself (because it is a line of symmetry of the square), and since \(\overline{AB}\parallel\overline{FG}\), the hexagon also maps onto itself.

• For option A: The perpendicular bisectors of the sides of the hexagon may not be lines of symmetry of the square. For example, if we take a perpendicular bisector of a side of the hexagon that is not aligned with the square's symmetry (due to \(\overline{AB}\parallel\overline{FG}\)), the square part of the combined figure will not map onto itself.
• For option B: The diagonals of the square are not lines of symmetry of the combined figure. If we reflect across a diagonal of the square, the hexagon (because of \(\overline{AB}\parallel\overline{FG}\)) will not map onto itself.
• For option D: This is incorrect as we have found lines of symmetry.

Answer:

C. either perpendicular bisector of the sides of the square