Question

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**#1.) determine the minimum number of degrees to rotate each regular polygon around its center to map it onto itself

#2.) sketch the equilateral triangle shown below after a 180° rotation. (place the sketch inside the box.)

after the rotation you drew above, would the triangle be mapped onto itself? why or why not?

**#3.) a regular polygon is rotated 288° around its center to be mapped onto itself. which polygon is it?

a. equilateral triangle
b. regular hexagon
c. regular pentagon
d. square

Explanation:

Step1: Formula for rotation angle

The formula to find the minimum rotation angle \(\theta\) for a regular \(n -\)sided polygon is \(\theta=\frac{360^{\circ}}{n}\)

For the pentagon (\(n = 5\))

\(\theta=\frac{360^{\circ}}{5}=72^{\circ}\)

For the non - agon (\(n = 9\))

\(\theta=\frac{360^{\circ}}{9} = 40^{\circ}\)

For the equilateral triangle (\(n=3\))

\(\theta=\frac{360^{\circ}}{3}=120^{\circ}\)

Step2: Analyzing the \(180^{\circ}\) rotation of an equilateral triangle

An equilateral triangle has \(n = 3\) sides. The minimum rotation angle is \(120^{\circ}\). Since \(180^{\circ}\) is not a multiple of \(120^{\circ}\) (\(180\div120 = 1.5\)), the triangle will not be mapped onto itself after a \(180^{\circ}\) rotation.

Step3: Solving for \(n\) in the rotation formula

We know that \(\theta=\frac{360^{\circ}}{n}\), and we are given that a rotation of \(288^{\circ}\) maps the polygon onto itself. Let \(k\) be a positive integer such that \(k\theta=288^{\circ}\) and \(\theta=\frac{360^{\circ}}{n}\). Then \(k\times\frac{360^{\circ}}{n}=288^{\circ}\), so \(n=\frac{360k}{288}=\frac{5k}{4}\)

Since \(n\) is a positive integer, when \(k = 4\), \(n = 5\)

Answer:

1. Pentagon: \(72^{\circ}\), Non - agon: \(40^{\circ}\), Equilateral triangle: \(120^{\circ}\)
2. The equilateral triangle is not mapped onto itself after a \(180^{\circ}\) rotation because \(180^{\circ}\) is not a multiple of its minimum rotation angle (\(120^{\circ}\))
3. c. Regular pentagon