1. Similarity and Transformation CHAPTER 7

2. Chapter 7 7.1 – SCALE DIAGRAMS AND ENLARGEMENTS 7.2 – SCALE DIAGRAMS AND REDUCTIONS 7.3 – SIMILAR POLYGONS

3. SCALE DIAGRAMS What is a scale diagram? Where are some places that we see enlargements in real life? Where are some places that we see reductions in real life?

4. SCALE FACTOR How can I tell what the scale has been changed by? We need to look at corresponding lengths, which are the matching lengths on the original diagram and the scale diagram. Using those lengths, we can find the scale factor, which is the proportion by which each length has been changed. Scale Factor = So, in this case: Scale factor = 5/2 = 2.5

5. EXAMPLE This photo of longhouses has dimensions 9 cm by 6 cm. The photo is to be enlarged by a scale factor of (7/2). Calculate the dimensions of the enlargements. If you have the scale factor, all you need to do is multiply it by the original to find the scale dimensions.  Conversely, if you have the scale diagram, all you need to do is divide it by the scale factor. The original is 9 by 6:  9 x (7/2) = 31.5 cm  6 x (7/2) = 21 cm The dimensions of the scale diagram is 31.5 cm by 21 cm.

6. EXAMPLE Scale diagrams always have equal angles.

7. SCALE FACTORS A scale factor is a proportion. What different ways can a proportion be represented? ½ 0.5 50% 1:2 fractions decimals ratios percents

8. SIMILAR POLYGONS Properties of Similar Polygons: When two polygons are similar: their corresponding angles are equal, and Their corresponding sides are proportional. It is also true that if two polygons have these properties, then the polygons are similar.

9. EXAMPLE Identify corresponding lengths: AB ≈ HG ≈ MN BC ≈ FG ≈ KM So, if they are similar, then they will have proportional scale factors.  AB/HG = BC/FG  8.5/8.4 = 2.5/2.4  1.01190… = 1.04166…  No, they aren’t similar.  AB/MN = BC/KM  8.5/5.25 = 2.5/1.5  1.619… = 1.666…  No, they aren’t similar.  HG/MN = FG/KM  8.4/5.25 = 2.4/1.5  1.6 = 1.6  They are similar! EFGH and JNKM are similar, since their corresponding lengths share the same proportion.

10. EXAMPLE a)AB/IJ = GH/QP  5.4/8.1 = GH/32.4  0.666… = GH/32.4  (32.4)(0.666…) = GH  GH = 21.6 m b) NP/FG = IJ/AB  NP/27 = 8.1/5.4  NP = 27(8.1/5.4)  NP = 40.5 m We need to use proportions.

11. Independent Practice P. 323-325, # 11, 12, 15. P. 329-330, # 6, 8, 15, 19. P. 341-343, # 4, 6, 9, 12, 13, 16.

12. Chapter 7 7.4 – SIMILAR TRIANGLES

13. GEOGEBRA Today, we’re going to the computer lab and will be using the program GeoGebra. It is important that once we get into the lab that you listen carefully to my instructions so that you can follow along with the process of the activity.

14. SIMILARITY

15. SIMILAR TRIANGLES Properties of Similar Triangles: To identify that ΔPQR and ΔSTU are similar, we only need to know that: ∠ P = ∠ S and ∠ Q = ∠ T and ∠ R = ∠ U or

16. EXAMPLE Since we have all the lengths, we can use corresponding lengths to check if the triangles are similar. The longest side has to go with the longest side, and so forth, so: SR ~ RQ TR ~ RP TS ~ QP Now we need to check the ratios: SR/RQ = 9/6 = 1.5 TR/RP = 7.5/5 = 1.5 TS/QP = 6/4 = 1.5 Since the corresponding sides are proportional, the triangles are similar.  ΔTSR ~ ΔPQR

17. EXAMPLE A surveyor wants to determine the width of a lake at two points on opposite sides of the lake. She measures distances and angles on land, then sketches this diagram. How can the surveyor determine the length HN to the nearest metre? We know from the diagram that: ∠ H = ∠ P ∠ N = ∠ Q ∠ J = ∠ J (the triangles share an angle) So, the triangles are similar! There corresponding sides must be proportional.  HJ/PJ = HN/PQ  515/210 = HN/230  2.45 = HN/230  (230)(2.45) = HN  HN = 564.0 m To the nearest metre, HN is 564 metres long.

18. TRY IT A surveyor used this scale diagram to determine the width of a river. The measurements he made and the equal angles are shown. What is the width, AB, to the nearest tenth of a metre? Which sides are corresponding lengths?  AC ~ CE (because it’s the length between ∠ C and the 90°angle for both triangles)  AB ~ DE So, now we can make our proportions:  AC/CE = AB/DE  28.9/73.2 = AB/98.3  (98.3)(28.9/73.2) = AB  AB = 38.8097… To the nearest metre, the river is 38.8 metres wide.

19. Chapter 7 PG. 349-351, # 5, 6, 9, 10, 12, 14, 15.

20. Chapter 7 7.5 – REFLECTIONS AND LINE SYMMETRY

21. SYMMETRY Look at each shape, and identify as many lines of symmetry as you can. If you need to, you can cut the shapes out. The colours need to be symmetric as well. None!

22. None

23. EXAMPLE Identify the triangles that are related to the red triangle by a line of reflection. Describe the position of each line of symmetry. The red triangle and triangle B are symmetric right from the red triangle’s base. You can see the triangle A is a mirror image of the red triangle, if it were flipped horizontally. Triangle D is a diagonal reflection of the red triangle. Triangle C does not have a line of symmetry with the red triangle, because it’s pointed in the other direction

24. EXAMPLE

25. Independent Practice PG. 357-359, # 4, 5, 7, 8, 10, 11.

26. Chapter 7 7.6 – ROTATIONS AND ROTATIONAL SYMMETRY 7.7 - IDENTIFYING TYPES OF SYMMETRY ON THE CARTESIAN PLANE

27. ROTATIONAL SYMMETRY A shape has rotational symmetry when it coincides with itself after a rotation of less than 360° about its centre. The number of times the shape coincides with itself, during a rotation of 360º, is the order of rotation. The shape above has rotational symmetry of order 4. We also talk about angle of rotation symmetry. Above, each rotation was 90 degrees, and it coincided with itself each time, so we say that its angle of rotation symmetry is 90 degrees.

28. EXAMPE

29. EXAMPLE (CONTINUED) a) We can see that it coincides with itself three times in 360º, so its order of rotation is 3.  360/3 = 120º is the angle of rotation symmetry. b) We can see that it coincides with itself twice in 360º, so its order of rotation is 2.  360/2 = 180º is the angle of rotation symmetry. c)The arrow does not coincide with itself when rotated 360º, so it does not have rotational symmetry. Try it.

30. EXAMPLE b) The resulting shape has rotational symmetry, of order 4 about point A.

31. CARTESIAN PLANE What’s a Cartesian plane? On the graph paper given to you, draw an x and a y axis. x y Trace the points A(1,3), B(3,1) and C(5,5). Then, join the points to form a triangle. A B C Do these three transformations, one at a time, using the original shape each time: a translate 2 units right, and 2 down. a rotation of 180° about vertex C a reflection in a line through AB Label and record the vertices of each new shape.

33. EXAMPLE For each pair of rectangles ABCD and EFGH, determine whether they are related by symmetry. We can see that there is no line that we can put down where one rectangle is the reflection of the other. So, let’s look for rotational symmetry.  We can see that ABCD is rotated 180º about point S and coincides with EFGH.  So, the rectangles have rotational symmetry of order 2 about S(0,3) S

34. EXAMPLE CONTINUED Try it! Remember, look for line symmetry and rotational symmetry. (This one has both). ABCD has corresponding points on EFGH that are equidistant from the x-axis.  So, the two rectangles are related by line symmetry, where the x-axis the line of symmetry. It also has rotational symmetry. What point could we rotate ABCD around to coincide with EFGH?  P(–2.5, 0)  It rotates 180º around point P(–2.5, 0)  So, the two rectangles are also related by rotational symmetry.

35. EXAMPLE For translations we sometimes use the notation where 4 units right is “R4” and 1 unit down would be “D1”

36. Independent Practice PG. 365-367, # 4, 8, 9, 10, 14, 15 PG. 373-375, # 3, 6, 7, 10, 12, 13, 15