Symmetry of polygons: lines of symmetry and rotations that map a shape onto itself
Grade 10 Math Worksheet

Circle the best answer.
Ari puts a rug on a grid. Which list gives every symmetry of the rectangle, with centers and mirror lines?
- Reflections across x = 3 and y = 2, and a 180° turn about (3, 2)
- Reflections across x = 3 and y = 2, with no turn
- Reflections across x = 3 and y = 2, and a 180° turn about (0, 0)
- Reflections across x = 3, y = 2 and both diagonals, and a 180° turn about (3, 2)
Bo draws a rhombus on a grid. Which list gives every symmetry, with centers and mirror lines?
Rhombus A(0, 0), B(4, 3), C(8, 0), D(4, -3). All four sides are 5 and the figure has no right angles. Rhombus A(0, 0), B(4, 3), C(8, 0), D(4, -3). All four sides are 5 and the figure has no right angles. −1 1 2 3 4 5 6 7 8 9 −4 −3 −2 −1 1 2 3 4 0 5 5 5 5 A B C D All four sides are 5 and the figure has no right angles. - Reflections across the lines through midpoints of opposite sides, and a 180° turn about (4, 0)
- Reflections across y = 0 and x = 4, with no turn
- Reflections across y = 0 and x = 4, and a 90° turn about (4, 0)
- Reflections across y = 0 and x = 4, and a 180° turn about (4, 0)
Esme draws a slanted parallelogram on a grid. Which list gives every symmetry, with centers and mirror lines?
Parallelogram A(0, 0), B(5, 0), C(7, 3), D(2, 3). The sides are 5 and √13, and no angle is a right angle. Parallelogram A(0, 0), B(5, 0), C(7, 3), D(2, 3). The sides are 5 and √13, and no angle is a right angle. −1 1 2 3 4 5 6 7 8 −1 1 2 3 4 0 5 √13 5 √13 A B C D The sides are 5 and √13, and no angle is a right angle. - No reflections; one 180° turn about (3.5, 1.5)
- No reflections; one 180° turn about (0, 0)
- No symmetry of any kind
- Reflections across x = 3.5 and y = 1.5, and a 180° turn about (3.5, 1.5)
Jun draws an isosceles trapezoid on a grid. Which list gives every symmetry, with centers and mirror lines?
Trapezoid A(0, 0), B(8, 0), C(6, 3), D(2, 3). AB ∥ DC, with AD = BC. Trapezoid A(0, 0), B(8, 0), C(6, 3), D(2, 3). AB ∥ DC, with AD = BC. −1 1 2 3 5 6 7 8 9 −1 1 2 3 4 0 A B C D AB ∥ DC, with AD = BC. - No symmetry of any kind
- Reflections across x = 4 and y = 1.5
- One reflection, across x = 4, and no turn symmetry
- One reflection, across x = 4, and a 180° turn about (4, 1.5)
AB = AD = √17. BD has slope -1. Kira thinks the kite ABCD is symmetric across the line y = x. Which computed fact is enough to show it?
Kite A(0, 0), B(1, 4), C(5, 5), D(4, 1). A and C lie on y = x. AB = AD = √17. BD has slope −1. Kite A(0, 0), B(1, 4), C(5, 5), D(4, 1). A and C lie on y = x. AB = AD = √17. BD has slope −1. y = x A(0, 0) B(1, 4) C(5, 5) D(4, 1) x y - AB = AD = √17, so the two sides at A are equal and the kite is symmetric
- A and C lie on y = x, so they stay put and the other corners must match too
- BD is perpendicular to AC, so y = x must be a mirror line for the kite
- BD has slope −1 with its midpoint (2.5, 2.5) on y = x, so y = x bisects BD at 90°
AB = CD = √26. AC = √65 and BD = √13. Luz thinks a half turn maps the quadrilateral ABCD onto itself. Which computed fact is enough to show it?
Quadrilateral A(0, 0), B(5, 1), C(7, 4), D(2, 3). AB = CD = √26. AC = √65 and BD = √13. Quadrilateral A(0, 0), B(5, 1), C(7, 4), D(2, 3). AB = CD = √26. AC = √65 and BD = √13. Quadrilateral A(0, 0), B(5, 1), C(7, 4), D(2, 3) x y A B C D - The diagonals are perpendicular, so the corners swap in pairs
- AC and BD have equal lengths, so the turn swaps the two diagonals
- The midpoint of AC and the midpoint of BD are the same point, (3.5, 2)
- AB = CD = √26, so opposite sides match under the turn, and that is enough
Pia proves the base angles of isosceles △ABC (AB = AC) are equal with a reflection across the bisector of ∠A. Which reason is best?
Isosceles triangle ABC with AB = AC and line ℓ bisecting ∠A. Isosceles triangle ABC with AB = AC and line ℓ bisecting ∠A. AB = AC l A B C - The reflection swaps B and C, so it maps ∠B onto ∠C, and reflections keep angles
- Base angles are equal in any triangle, because the base is one of its sides
- AB = AC, so ∠B = ∠C by the definition of an isosceles triangle with equal sides
- The reflection keeps B fixed, so ∠B stays the same and ∠C is unchanged too
In isosceles △ABC (AB = AC), the reflection across ℓ swaps B and C. Ren asks why ℓ is perpendicular to BC and crosses its midpoint. What is the reason?
Isosceles triangle ABC with AB = AC. A reflection across the line ℓ through A swaps B and C. Isosceles triangle ABC with AB = AC. A reflection across the line ℓ through A swaps B and C. AB = AC l A B C - An angle bisector is perpendicular to the opposite side, in any triangle you pick
- A reflection that swaps B and C has the perpendicular bisector of BC as mirror line
- ℓ passes through A, so it must pass through the midpoint of BC as well
- BC is parallel to ℓ, so ℓ splits BC in half at the same level as the base
Tala says an equilateral triangle maps onto itself by a 120° turn about its center, sending A to B to C. Why must each angle be 60°?
Equilateral triangle ABC. A 120° turn about its center sends A to B, B to C and C to A. Equilateral triangle ABC. A 120° turn about its center sends A to B, B to C and C to A. 120° A B C center - 120° is the turn, so each angle is 120° and the three angles add to 360°
- A reflection keeps A fixed, so ∠A = 60° and the other two angles follow
- The turn sends ∠A to ∠B to ∠C, so the three angles are equal, and add to 180°
- The three sides are equal, so each angle is 60° by the definition of an equilateral triangle
Zee proves the diagonals of a rectangle ABCD are equal, using the reflection across its vertical midline. Which reason is best?
Rectangle ABCD with A top left, B top right, C bottom right, D bottom left. The vertical midline swaps A with B and D with C. Rectangle ABCD with A top left, B top right, C bottom right, D bottom left. The vertical midline swaps A with B and D with C. vertical midline A B C D - The diagonals bisect each other, so they must be equal in length, as in a rectangle
- A half turn about the center maps AC onto itself, so AC must equal BD
- The reflection swaps A with B and C with D, so AC maps onto BD, and length is kept
- The diagonals of any quadrilateral are equal, so these two diagonals are equal