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

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

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

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

Similar Polygons: Corresponding Parts and Proportions B 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)

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

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

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 = 20 12 = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖

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 = 20 12 = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖 𝑅𝑇 𝑋𝑍 = 25 15 = 5 3

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 = 20 12 = 5 3 𝑿 𝒁 𝒀 𝟏𝟓 𝟏𝟐 𝟏𝟖 𝑅𝑇 𝑋𝑍 = 25 15 = 5 3 𝑆𝑇 𝑌𝑍 = 30 18 = 5 3

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

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

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

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

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 = 20 25 = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓

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 = 20 25 = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5

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 = 20 25 = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5 𝐻𝐽 𝑋𝑊 = 12 15 = 4 5

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 = 20 25 = 4 5 𝒁 𝑾 𝑿 𝒀 𝟐𝟓 𝟐𝟎 𝟑𝟎 𝟏𝟓 𝐺𝐻 𝑌𝑋 = 24 30 = 4 5 𝐻𝐽 𝑋𝑊 = 12 15 = 4 5 𝐽𝐹 𝑊𝑍 = 16 20 = 4 5

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

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

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

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

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

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

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

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

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

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

Example 7 In the diagram, ∆𝑨𝑩𝑪 ~ ∆𝑮𝑯𝑱 c.) If the perimeter of ABC For Perimeter of ABC 𝐴𝐵𝐶=12+18+24 𝑨𝑩𝑪=𝟓𝟒 𝑩 𝟏𝟐 𝑨 𝟏𝟖 𝑪 𝒙 𝑯 𝟏𝟔 𝑮 𝟏𝟐 𝑱 𝟖

Example 7 In the diagram, ∆𝑨𝑩𝑪 ~ 𝑮𝑯𝑱 c.) If the perimeter of ABC For Perimeter of ABC 𝐺𝐻𝐽=8+12+16 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

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

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

Example 8 In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 b.) Find the value of x. 𝑥=13 𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

Example 8 In the diagram, 𝑨𝑩𝑪𝑫𝑬 ~ 𝑭𝑮𝑯𝑱𝑲 c.) If the perimeter of ABCDE 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝐴𝐵𝐶𝐷𝐸 =10+13+10+9+8=50 𝑭 𝟐𝟎 𝑯 𝑮 𝑲 𝑱 𝟏𝟔 𝟏𝟖 𝟐𝟔 𝑨 𝟏𝟎 𝑪 𝑩 𝑬 𝑫 𝒙

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

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