1. 6-1: Use Similar Polygons
Geometry Chapter 6
6-1: Use Similar Polygons

2. Warm-Up The given triangles are congruent.
1.) List the corresponding sides
2.) List the corresponding angles
3.) Give a congruence statement for the congruent triangles

3. Use Similar Polygons
Objective: Students will be able to use proportions and corresponding parts to identify and solve similar polygons.
Agenda
Similar Polygons
Scale Factors
Perimeters of Similar Polygons

4. Definition: Similar Polygons
Two polygons are similar if their vertices can be paired so that:
1.) Corresponding angles are congruent
2.) Corresponding sides are in proportion (i.e. Their side lengths have the same ratio.) A B C D
Example: E F G H

5. Similar Polygons: Corresponding Parts and ProportionsB C D
Corresponding Angles:
<𝐴≅ <𝐸
<𝐵≅ <𝐹
<𝐶≅ <𝐺
<𝐷≅ <𝐻 E F G H
Proportion of Corresponding Sides:
𝐴𝐵 𝐸𝐹 = 𝐵𝐶 𝐹𝐺 = 𝐶𝐷 𝐺𝐻 = 𝐷𝐴 𝐻𝐸 =𝑆𝑐𝑎𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟(𝑘)

6. Example 1 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖 In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
a.) List all pairs of congruent angles
b.) Check that the ratios of corresponding side lengths are equal
c.) Write the ratios of the corresponding side lengths in a statement of proportionality 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖

7. Example 1 In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
a.) List all pairs of congruent angles
Corresponding Angles:
<𝑅≅ <𝑋
<𝑆≅ <𝑌
<𝑇≅ <𝑍 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖

8. Example 1 In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
b.) Check that the ratios of corresponding side lengths are equal 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎
𝑅𝑆 𝑋𝑌 = = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖

9. Example 1 In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
b.) Check that the ratios of corresponding side lengths are equal 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎
𝑅𝑆 𝑋𝑌 = = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖
𝑅𝑇 𝑋𝑍 = = 5 3

10. Example 1 In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
b.) Check that the ratios of corresponding side lengths are equal 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎
𝑅𝑆 𝑋𝑌 = = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖
𝑅𝑇 𝑋𝑍 = = 5 3
𝑆𝑇 𝑌𝑍 = = 5 3

11. Example 1 𝑻 𝑅𝑆 𝑋𝑌 = 𝑅𝑇 𝑋𝑍 = 𝑆𝑇 𝑌𝑍 𝒁 𝟑𝟎 𝟐𝟓 𝟏𝟖 𝟏𝟓 𝑆𝑐𝑎𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟(𝑘)= 5 3 𝑹
In the diagram, ∆𝑹𝑺𝑻~∆𝑿𝒀𝒁
c.) Write the ratios of the corresponding side lengths in a statement of proportionality 𝑹 𝑻 𝑺 𝟐𝟓 𝟐𝟎 𝟑𝟎
𝑅𝑆 𝑋𝑌 = 𝑅𝑇 𝑋𝑍 = 𝑆𝑇 𝑌𝑍 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖
𝑆𝑐𝑎𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟(𝑘)= 5 3

12. Example 2 𝑲 𝑱 𝑳 𝟑𝟐 𝟐𝟖 𝑸 𝑷 𝑹 𝟐𝟒 𝟐𝟏 In the diagram, ∆𝑱𝑲𝑳~∆𝑷𝑸𝑹
a.) List all pairs of congruent angles
b.) Check that the ratios of corresponding side lengths are equal
c.) Write the ratios of the corresponding side lengths in a statement of proportionality 𝑲 𝑱 𝑳 𝟑𝟐 𝟐𝟖 𝑸 𝑷 𝑹 𝟐𝟒 𝟐𝟏

13. Example 3
For the given polygons,
a.) Determine whether the polygons are similar
b.) If the polygons are similar, write a similarity statement and find the scale factor. 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓

14. Example 3
a.) To determine if the polygons are similar, you must show that all the corresponding side lengths are proportional 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐

15. Example 3 𝟏𝟔 𝑭 𝑱 𝐹𝐺 𝑍𝑌 = 20 25 = 4 5 𝟐𝟎 𝟏𝟐 𝟐𝟎 𝒁 𝑾 𝑯 𝟐𝟒 𝑮 𝟏𝟓 𝟐𝟓 𝑿 𝟑𝟎 𝒀
From the corresponding angles, the side proportions are as follows 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐
𝐹𝐺 𝑍𝑌 = = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓

16. Example 3 𝟏𝟔 𝑭 𝑱 𝐹𝐺 𝑍𝑌 = 20 25 = 4 5 𝟐𝟎 𝟏𝟐 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5 𝟐𝟎 𝒁 𝑾
From the corresponding angles, the side proportions are as follows 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐
𝐹𝐺 𝑍𝑌 = = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓
𝐺𝐻 𝑌𝑋 = = 4 5

17. Example 3 𝟏𝟔 𝑭 𝑱 𝐹𝐺 𝑍𝑌 = 20 25 = 4 5 𝟐𝟎 𝟏𝟐 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5 𝟐𝟎 𝒁 𝑾
From the corresponding angles, the side proportions are as follows 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐
𝐹𝐺 𝑍𝑌 = = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓
𝐺𝐻 𝑌𝑋 = = 4 5
𝐻𝐽 𝑋𝑊 = = 4 5

18. Example 3 𝟏𝟔 𝑭 𝑱 𝐹𝐺 𝑍𝑌 = 20 25 = 4 5 𝟐𝟎 𝟏𝟐 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5 𝟐𝟎 𝒁 𝑾
From the corresponding angles, the side proportions are as follows 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐
𝐹𝐺 𝑍𝑌 = = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓
𝐺𝐻 𝑌𝑋 = = 4 5
𝐻𝐽 𝑋𝑊 = = 4 5
𝐽𝐹 𝑊𝑍 = = 4 5

19. Example 3
b.) As seen in part (a), all the proportions were equal, so the polygons are similar with the statement 𝑭𝑱𝑮𝑯 ~ 𝒁𝒀𝑿𝑾 With scale factor k= 𝟒 𝟓 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓 𝑭 𝟐𝟎 𝑱 𝑯 𝑮 𝟏𝟔 𝟐𝟒 𝟏𝟐

20. Example 4
For the given polygons,
a.) Determine whether the polygons are similar
b.) If the polygons are similar, write a similarity statement and find the scale factor. 𝟑𝟎 𝟐𝟖 𝟑𝟑 𝟏𝟐 𝑹 𝑺 𝑼 𝑻 𝟐𝟎 𝟏𝟖 𝟐𝟐 𝟖 𝑫 𝑬 𝑮 𝑭

21. Example 5
In the diagram, ∆𝑫𝑬𝑭 ~ ∆𝑴𝑵𝑷. Find the value of x. 𝑴 𝑵 𝑷 𝟏𝟔 𝟏𝟐 𝟐𝟎 𝑫 𝑬 𝑭 𝟗 𝟏𝟐 𝒙

22. Example 5 𝑴 𝑵 𝑷 𝟏𝟔 𝟏𝟐 𝟐𝟎 𝑀𝑁 𝐷𝐸 = 𝑁𝑃 𝐸𝐹 𝑫 𝑬 𝑭 𝟗 𝟏𝟐 𝒙 12 9 = 20 𝑥
To solve for x, make a proportion that involves a pair of corresponding sides with known values, along with the pair that has the x. 𝑴 𝑵 𝑷 𝟏𝟔 𝟏𝟐 𝟐𝟎
𝑀𝑁 𝐷𝐸 = 𝑁𝑃 𝐸𝐹 𝑫 𝑬 𝑭 𝟗 𝟏𝟐 𝒙
12 9 = 20 𝑥
12𝑥=180
𝒙=𝟏𝟓

23. Example 6
In the diagram, 𝑨𝑩𝑪𝑫 ~ 𝑸𝑹𝑺𝑻. Find the value of x. 𝑨 𝟏𝟎 𝑩 𝑪 𝑫 𝟏𝟐 𝟏𝟔 𝒙 𝑸 𝟓 𝑹 𝑺 𝑻 𝟔 𝟖 𝟒

24. Example 6 𝐴𝐵 𝑄𝑅 = 𝐵𝐶 𝑅𝑆 𝑨 𝟏𝟐 𝑩 𝑸 𝟔 𝑹 12 6 = 𝑥 4 𝒙 𝟏𝟎 𝟓 𝟒 6𝑥=48 𝑻 𝑺 𝟖 𝑪
In the diagram, 𝑨𝑩𝑪𝑫 ~ 𝑸𝑹𝑺𝑻. Find the value of x.
𝐴𝐵 𝑄𝑅 = 𝐵𝐶 𝑅𝑆 𝑨 𝟏𝟎 𝑩 𝑪 𝑫 𝟏𝟐 𝟏𝟔 𝒙 𝑸 𝟓 𝑹 𝑺 𝑻 𝟔 𝟖 𝟒
12 6 = 𝑥 4
6𝑥=48
𝒙=𝟖

25. 𝐴𝐵+𝐵𝐶+𝐶𝐷+𝐷𝐴 𝐸𝐹+𝐹𝐺+𝐺𝐻+𝐻𝐸 = 𝐴𝐵 𝐸𝐹 = 𝐵𝐶 𝐹𝐺 = 𝐶𝐷 𝐺𝐻 = 𝐷𝐴 𝐻𝐸
Theorem 6,1
Theorem 6.1: If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths. A B C D E F G H
If 𝐴𝐵𝐶𝐷 ~ 𝐸𝐹𝐺𝐻, then
𝐴𝐵+𝐵𝐶+𝐶𝐷+𝐷𝐴 𝐸𝐹+𝐹𝐺+𝐺𝐻+𝐻𝐸 = 𝐴𝐵 𝐸𝐹 = 𝐵𝐶 𝐹𝐺 = 𝐶𝐷 𝐺𝐻 = 𝐷𝐴 𝐻𝐸

26. Example 7
In the diagram, ∆𝑨𝑩𝑪 ~ 𝑮𝑯𝑱 a.) Find the scale factor from ABC to GHJ b.) Find the value of x c.) Find the perimeter of ABC 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖

27. Example 7
In the diagram, ∆𝑨𝑩𝑪 ~ 𝑮𝑯𝑱 a.) Find the scale factor from ABC to GHJ 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙
Scale Factor
𝐴𝐵 𝐺𝐻 = 12 8 = 𝟑 𝟐 =𝒌 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖

28. Example 7 In the diagram, ∆𝑨𝑩𝑪 ~ 𝑮𝑯𝑱 b.) Find the value of x For x
𝐴𝐵 𝐺𝐻 = 𝐵𝐶 𝐻𝐽
3 2 = 𝑥 16
2𝑥=48
𝒙=𝟐𝟒 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖

29. Example 7 In the diagram, ∆𝑨𝑩𝑪 ~ ∆𝑮𝑯𝑱 c.) If the perimeter of ABC
For Perimeter of ABC
𝐴𝐵𝐶=
𝑨𝑩𝑪=𝟓𝟒 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖

30. Example 7 In the diagram, ∆𝑨𝑩𝑪 ~ 𝑮𝑯𝑱 c.) If the perimeter of ABC
For Perimeter of ABC
𝐺𝐻𝐽=
𝐺𝐻𝐽=36
Then
𝐴𝐵𝐶 𝐺𝐻𝐽 = 3 2
𝐴𝐵𝐶 36 = 3 2
𝑨𝑩𝑪=𝟓𝟒 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙

31. Example 8
In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 a.) Find the scale factor from FGHJK to ABCDE b.) Find the value of x. c.) If the perimeter of ABCDE 𝑭 𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

32. Example 8
In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 a.) Find the scale factor from FGHJK to ABCDE 𝑘= =2 𝑭 𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

33. Example 8 In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 b.) Find the value of x. 𝑥=13𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

34. Example 8
In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 c.) If the perimeter of ABCDE 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝐴𝐵𝐶𝐷𝐸 = =50 𝑭 𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

35. Extra Definition: Altitude
An altitude of a triangle is the perpendicular segment from a vertex to the opposite side, or to the line that contains the opposite side. 𝑨 𝑩 𝑪
Altitude from B to 𝑨𝑪 𝑩 𝑨 𝑪

36. Extra Definition: Median
A median of a triangle is a segment from a vertex to the midpoint of the opposite side. 𝑨 𝑩 𝑪
Centroid