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Warm-up Use Geometer’s Sketchpad to construct 2 non-overlapping, similar triangles whose corresponding side lengths are in the ratio of 3 to 2, and display.

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Presentation on theme: "Warm-up Use Geometer’s Sketchpad to construct 2 non-overlapping, similar triangles whose corresponding side lengths are in the ratio of 3 to 2, and display."— Presentation transcript:

1 Warm-up Use Geometer’s Sketchpad to construct 2 non-overlapping, similar triangles whose corresponding side lengths are in the ratio of 3 to 2, and display the perimeter of each triangle.

2 In questions 1 – 4, identify whether the triangles can be proven similar, and state the similarity theorem that supports the conclusion. 6 4 3 4.5 A B C X Y Z 1. 2. 3. 4. Yes, AA~ Yes, SAS~ Yes, SSS~ Yes, AA~

3 5.In the diagram below, rectangle RSTU is dilated (center P) and its image is rectangle RSTU. What is the scale factor of this dilation? 1:3

4 6.A blueprint of a triangular garden with sides measuring 2, 6, and 7 inches is placed on an overhead projector and projected onto a screen, as shown below. The shortest side of the triangle on the screen is 10 inches. How many inches is the longest side of the triangle on the screen? 35 inches

5 1.  A   B (Isosceles Triangle Theorem) 2.  PMA and  PNM are congruent right angles. 3.  AMP   BNP (AA  ) 8 CN = CB – BN= CA – BN = CM + AM – BN= CM + AM – 2AM = CM – AM

6 Prove: In the regular pentagon,  ADB =  CAD = Therefore,  ADB   APD (AA~) (Definition of similar triangles) 8.In the last homework assignment, you conjectured that in the regular pentagon shown, the ratio of AB to AD is equal to the ratio of AD to AP. Prove this conjecture. 36 A P D A D B Since the pentagon is regular, making  ABD isosceles, so  BAD =  ABD = 36  36 Similarly,  ACD isosceles, so  ACD =  ADC = 36  36

7 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD 

8 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD  

9 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD   

10 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD    70  35  55  20 

11 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD    

12 Which of the following triangles are similar to  ABC (there may be more than one answer). a.  AFE b.  DBE c.  DFC d.  AED e.  GFC f.  ABD    

13 Recall that the median of a trapezoid connects the midpoints of the non-parallel sides and its length is the average of the lengths of the bases. Is it possible for the diagonals of an isosceles trapezoid to be the same length as the median of the trapezoid? Explain. Suppose AC = MN Since the diagonals of an isosceles trapezoid are congruent, AC + BD = 2(AC). So AB + DC = (AP + PC) + (BP + PD) Contradiction Since we know, then AB + DC = 2(MN) =2(AC). > AB > DC by the triangle inequality (the sum of the lengths of any two sides of a triangle is greater than the length of the third side.) A B C D M N P = AP + BP + PC + PD 10. Therefore, AB + DC = AC + BD It is not possible.

14

15 AREA

16 The ratio of the perimeters of 2 similar triangles is equal to the scale factor. The ratio of the areas of 2 similar triangles is equal to the square of the scale factor.

17 b h L W Area of a rectangle A = bh

18 b h Area of a parallelogram, rhombus, rectangle, and square: A = bh b b h Area of a rectangle A = bh

19 h b Area of a triangle: A =bh h b b h L1L1 L2L2 A = ½ L 1 L 2

20 Not What it Seems An Area Paradox

21 What is the area of the right triangle shown? A = ½(13)(5) A = 32.5 5 13

22 A = ½(13)(5) A = 32.5

23 A = ½(13)(5) A = 32.5

24 A = ½(13)(5) A = 32.5

25 A = ½(13)(5) A = 32.5

26 What’s the area now? A = ½(13)(5) A = 32.5 A = 32.5–1 = 31.5

27 A = 64 What is the area of the square shown?

28 A = 64

29

30

31 How is this possible? What’s the area now? Area = 65 A = 64

32 h Area of a trapezoid: A = h 1 2 (b 1 + b 2 )

33 h Area of a trapezoid: A = h A = (median)(altitude) median 1 2 (b 1 + b 2 )

34 Practice: 1.A square has the same area as the parallelogram shown. What is the perimeter of the square? 16 inches 9 inches A B C M 2.CM is a median of  ABC. Explain why the areas of  AMC and  BMC must be equal. 3.The lengths of the bases of a certain trapezoid are 13 and 10. If the area of the trapezoid is 92, what is the length of the altitude of the trapezoid? 48 inches 8 inches Theorem: The median of a triangle divides the triangle in to triangles of equal area.


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