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SAS

SAS. Subtraction. 180. 270. Chapter 7 Section 7.3. Rotations. PQR is Rotated about point P to form PQ’R’ P is the center of rotation QPQ’ or RPR’ is the angle of rotation. Theorem 7.2 Rotation Theorem A rotation is an isometry.

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SAS

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  1. SAS Subtraction 180 270

  2. Chapter 7Section 7.3 Rotations

  3. PQR is Rotated about point P to form PQ’R’ P is the center of rotation QPQ’ or RPR’ is the angle of rotation Theorem 7.2Rotation Theorem A rotation is an isometry

  4. A 120º rotation of an equilateral triangle shows that it has rotational symmetry

  5. 45° Clockwise Rotation 90° Clockwise Rotation Yes the figure has rotational symmetry a 90° rotation

  6. 45° Clockwise Rotation 90° Clockwise Rotation 120° Clockwise Rotation Yes the figure has rotational symmetry a 120° rotation

  7. 45° Clockwise Rotation Yes the figure has rotational symmetry a 45° rotation

  8. 45° Clockwise Rotation Yes the figure has rotational symmetry a 45° rotation

  9. HPA GPH

  10. Theorem P ABC  A’B’C’ by a reflection over m A’B’C’  A”B”C” by a reflection over k ABC  A”B”C” by a rotation around P

  11. Theorem 7.3 Continued P ABC  A”B”C” by a rotation around P Angle of rotation = 2(40) = 80

  12. Angle of Rotation = 2(32) = 64 Angle of Rotation = 2x = 128 x = 64

  13. Rotate your paper the given rotation and read off the points like you did not rotate it. A(1, 3) B(3, 3) C(3, -1) D(1, -1)

  14. Rotate your paper the given rotation and read off the points like you did not rotate it. A(2, 0) B(3, -2) C(2, -4) D(1, -2)

  15. Rotate your paper the given rotation and read off the points like you did not rotate it. A(-2, 0) B(-4, 1) C(-4, 5) D(-2, 4)

  16. HW #3Pg 416-419 13-38, 43

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