Question

which of the following transformations carry this regular polygon onto itself? image of a regular triangle with a line of reflection l rotation of 90° counterclockwise; rotation of 120° counterclockwise; rotation of 90° clockwise; reflection across l

Explanation:

Step1: Analyze the regular polygon

The figure is a regular triangle (equilateral triangle). For a regular \( n \)-sided polygon, the angle of rotational symmetry is \( \frac{360^\circ}{n} \). For \( n = 3 \), the rotational symmetry angle is \( \frac{360^\circ}{3}=120^\circ \). So a rotation of \( 120^\circ \) (clockwise or counter - clockwise) will map the equilateral triangle onto itself. Also, an equilateral triangle has 3 lines of symmetry. The line \( l \) in the figure is a line of symmetry (it passes through a vertex and the mid - point of the opposite side), so reflection across \( l \) will map the triangle onto itself. Rotations of \( 90^\circ \) (clockwise or counter - clockwise) do not map an equilateral triangle onto itself because \( 90^\circ\) is not a multiple of \( 120^\circ \).

Step2: Evaluate each transformation

• Rotation of \( 90^\circ \) counter - clockwise: As \( 90^\circ

eq120^\circ, 240^\circ, 360^\circ \) (the angles of rotational symmetry for an equilateral triangle), this rotation does not carry the polygon onto itself.

• Rotation of \( 120^\circ \) counter - clockwise: Since \( 120^\circ=\frac{360^\circ}{3} \), this rotation will carry the equilateral triangle onto itself.
• Rotation of \( 90^\circ \) clockwise: Similar to the counter - clockwise \( 90^\circ \) rotation, \( 90^\circ\) is not a rotational symmetry angle of the equilateral triangle, so it does not carry the polygon onto itself.
• Reflection across \( l \): Since \( l \) is a line of symmetry of the equilateral triangle, reflecting across \( l \) will carry the triangle onto itself.

Answer:

The transformations that carry the regular polygon (equilateral triangle) onto itself are "rotation of \( 120^\circ \) counterclockwise" and "reflection across \( l \)".