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, with no turn
- Reflections across x = 3 and y = 2, and a 180° turn about (0, 0)
- Reflections across x = 3 and y = 2, and a 180° turn about (3, 2)
- 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 y = 0 and x = 4, with no turn
- Reflections across the lines through midpoints of opposite sides, and a 180° turn about (4, 0)
- 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. - Reflections across x = 3.5 and y = 1.5, and a 180° turn about (3.5, 1.5)
- No reflections; one 180° turn about (3.5, 1.5)
- No reflections; one 180° turn about (0, 0)
- No symmetry of any kind
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. - One reflection, across x = 4, and a 180° turn about (4, 1.5)
- One reflection, across x = 4, and no turn symmetry
- Reflections across x = 4 and y = 1.5
- No symmetry of any kind
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
Milo thinks the triangle with corners (0, 0), (6, 0) and (3, 4) is symmetric across x = 3. Which computed fact is enough?
Triangle A(0, 0), B(6, 0), C(3, 4). The sides are AB = 6 and AC = BC = 5. Triangle A(0, 0), B(6, 0), C(3, 4). The sides are AB = 6 and AC = BC = 5. −1 1 2 3 4 5 6 7 −1 1 2 3 4 5 0 AB = 6 AC = 5 BC = 5 A B C - The base has length 6, so each half of the base has length 3
- The triangle has a right angle at C, so its legs are mirror images
- C(3, 4) is above the base, so the picture looks balanced and symmetric
- x = 3 is the perpendicular bisector of AB, and C(3, 4) lies on x = 3
Noah wonders whether the triangle with corners (0, 0), (6, 0) and (2, 5) has any line of symmetry. What is the best answer?
Triangle A(0, 0), B(6, 0), C(2, 5). AB = 6, AC = √29, BC = √41. Triangle A(0, 0), B(6, 0), C(2, 5). AB = 6, AC = √29, BC = √41. −1 1 2 3 4 5 6 7 −1 1 2 3 4 5 6 0 AB = 6 AC = √29 BC = √41 A B C - Yes: x = 3 is the perpendicular bisector of the base AB, so it is a line of symmetry
- Yes: every triangle has a line of symmetry through its longest side
- No: the triangle has a vertex at (2, 5), which is not on the base
- No: the sides are all different, and a reflection must swap two equal sides
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 keeps B fixed, so ∠B stays the same and ∠C is unchanged too
- AB = AC, so ∠B = ∠C by the definition of an isosceles triangle with equal sides
- Base angles are equal in any triangle, because the base is one of its sides
- The reflection swaps B and C, so it maps ∠B onto ∠C, and reflections keep angles
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 - BC is parallel to ℓ, so ℓ splits BC in half at the same level as the base
- ℓ passes through A, so it must pass through the midpoint of BC as well
- 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
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 - A half turn about the center maps AC onto itself, so AC must equal BD
- The diagonals bisect each other, so they must be equal in length, as in a rectangle
- The diagonals of any quadrilateral are equal, so these two diagonals are equal
- The reflection swaps A with B and C with D, so AC maps onto BD, and length is kept