Question

which of the following transformations carry this regular polygon onto itself?rotation of 60° counterclockwiserotation of 72° counterclockwiserotation of 120° counterclockwiserotation of 90° clockwise

Explanation:

Step1: Calculate the rotational symmetry angle of a regular pentagon

The formula for the rotational symmetry angle of a regular \(n -\)sided polygon is \(\frac{360^{\circ}}{n}\). For a pentagon (\(n = 5\)), \(\frac{360^{\circ}}{5}=72^{\circ}\). A rotation of \(k\times72^{\circ}\) (\(k = 1,2,3,\cdots\)) will map the regular pentagon onto itself.

Step2: Analyze each option

• For a rotation of \(60^{\circ}\): Since \(60^{\circ}\) is not a multiple of \(72^{\circ}\), a rotation of \(60^{\circ}\) will not map the regular pentagon onto itself.
• For a rotation of \(72^{\circ}\): Using the formula \(\frac{360^{\circ}}{n}\) with \(n = 5\), a rotation of \(72^{\circ}\) (where \(k = 1\) in \(k\times72^{\circ}\)) will map the regular pentagon onto itself.
• For a rotation of \(120^{\circ}\): Since \(120^{\circ}\) is not a multiple of \(72^{\circ}\) (\(\frac{120}{72}=\frac{5}{3}\), not an integer), a rotation of \(120^{\circ}\) will not map the regular pentagon onto itself.
• For a rotation of \(90^{\circ}\): Since \(90^{\circ}\) is not a multiple of \(72^{\circ}\) (\(\frac{90}{72}=\frac{5}{4}\), not an integer), a rotation of \(90^{\circ}\) will not map the regular pentagon onto itself.

Answer:

rotation of \(72^{\circ}\) counterclockwise