Question

which of the following transformations carry this regular polygon onto itself? image of a square with a line l rotation of 30° clockwise; rotation of 90° clockwise; reflection across l; rotation of 90° counterclockwise

Explanation:

Step1: Analyze the regular polygon (square)

A square has rotational symmetry of order 4, meaning the smallest angle of rotation that maps it onto itself is \( \frac{360^\circ}{4}=90^\circ \). So rotations of \( 90^\circ \) (clockwise or counterclockwise) will carry it onto itself. A \( 30^\circ \) rotation is not a multiple of \( 90^\circ \), so it won't.

Step2: Analyze reflection

The line \( l \) is a diagonal of the square (from top - left to bottom - middle? Wait, looking at the square, the line \( l \) seems to be a line of symmetry (maybe a diagonal or a line through mid - points? Wait, in a square, reflections across its diagonals or lines through mid - sides are lines of symmetry. If \( l \) is a line of symmetry (like a diagonal or a vertical/horizontal line or a line through mid - points), then reflection across \( l \) will carry the square onto itself. But the \( 30^\circ \) rotation is incorrect. So the correct transformations are rotation of \( 90^\circ \) clockwise, rotation of \( 90^\circ \) counterclockwise, and reflection across \( l \) (assuming \( l \) is a line of symmetry). But the wrong one is "rotation of \( 30^\circ \) clockwise". Wait, the question is which transformations carry it onto itself. Let's re - evaluate:

• Rotation of \( 30^\circ \) clockwise: A square's rotational symmetry angles are multiples of \( 90^\circ \) (\( 90^\circ, 180^\circ, 270^\circ, 360^\circ \)). \( 30^\circ \) is not a multiple of \( 90^\circ \), so this rotation does not carry the square onto itself.
• Rotation of \( 90^\circ \) clockwise: Since \( 90^\circ \) is a rotational symmetry angle of the square, this rotation carries the square onto itself.
• Reflection across \( l \): If \( l \) is a line of symmetry (e.g., a diagonal, a line through mid - sides), then reflection across \( l \) will map the square onto itself.
• Rotation of \( 90^\circ \) counterclockwise: Similar to \( 90^\circ \) clockwise, this is a rotational symmetry angle, so it carries the square onto itself.

But the original marked options have a mistake (the \( 30^\circ \) rotation is marked but it's wrong). Assuming we are to find the correct ones:

Step3: Correcting the choices

The correct transformations are:

• Rotation of \( 90^\circ \) clockwise: Because \( 90^\circ \) is the angle of rotational symmetry for a square (\( \frac{360^\circ}{4}=90^\circ \)).
• Rotation of \( 90^\circ \) counterclockwise: Same as above, \( 90^\circ \) rotation in the counterclockwise direction also maps the square onto itself.
• Reflection across \( l \): If \( l \) is a line of symmetry (like a diagonal or a line through mid - points of opposite sides) of the square, then reflecting across \( l \) will map the square onto itself.

The incorrect one is "rotation of \( 30^\circ \) clockwise" because \( 30^\circ \) is not a rotational symmetry angle of the square.

But if we are to list the correct ones:

Answer:

• Rotation of \( 90^\circ \) clockwise
• Rotation of \( 90^\circ \) counterclockwise
• Reflection across \( l \)

(Note: The "rotation of \( 30^\circ \) clockwise" is incorrect and should not be selected. If we assume the line \( l \) is a line of symmetry for the square, then reflection across \( l \) is correct. The \( 90^\circ \) rotations are correct because the square has rotational symmetry of order 4, so \( 90^\circ \) rotations (clockwise or counterclockwise) map it onto itself.)