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, y = 2 and both diagonals, and a 180° turn about (3, 2)
- Reflections across x = 3 and y = 2, and a 180° turn about (0, 0)
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, and a 90° turn about (4, 0)
- Reflections across y = 0 and x = 4, and a 180° turn about (4, 0)
- Reflections across y = 0 and x = 4, with no turn
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. - Reflections across x = 4 and y = 1.5
- No symmetry of any kind
- 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 - BD has slope −1 with its midpoint (2.5, 2.5) on y = x, so y = x bisects BD at 90°
- A and C lie on y = x, so they stay put and the other corners must match too
- AB = AD = √17, so the two sides at A are equal and the kite is symmetric
- BD is perpendicular to AC, so y = x must be a mirror line for the kite
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 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
- AC and BD have equal lengths, so the turn swaps the two diagonals
- The diagonals are perpendicular, so the corners swap in pairs
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: 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
- Yes: x = 3 is the perpendicular bisector of the base AB, so it is a line of symmetry
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
- 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
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 - The three sides are equal, so each angle is 60° by the definition of an equilateral triangle
- 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°
- 120° is the turn, so each angle is 120° and the three angles add to 360°
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 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
- The diagonals bisect each other, so they must be equal in length, as in a rectangle