2018 Geometri Transformasi Perkalian 5 Isometri Kelompok 5 Rombel 3
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Tugas Akhir Mata Kuliah Geometri Transformasi FMIPA Unnes Rombel 03 Kelompok 5 Yosia Adi Setiawan (4101415010) Devy Kurniawati (4101415024) Julia Dwi P. (4101415042) Siti Jami'atun (4101415067) Azizah Galuh P. (4101415120) Info: yosia010@students.unnes.ac.id
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2018 Geometri Transformasi Perkalian 5 Isometri Kelompok 5 Rombel 3
- 1. Komposisi 5 Isometri Dasar (Geometri Transformasi) MATHEMATICS DEPARTMENT MATHEMATICS AND NATURAL SCIENCE FACULTY UNIVERSITAS NEGERI SEMARANG 2018 • Lecture: Drs. Suhito, M.Pd
- 2. 1 Geometri Transfomasi Nama Kelompok : • Yosia Adi Setiawan (4101415010) • Devy Kurniawati (4101415024) • Julia Dwi Pritanti (4101415042) • Siti Jami’atun (4101415067) • Azizah Galuh Puspasari (4101415120)
- 3. . Tugas Kelompok Geometri Transformasi Bukti langsung dan Bukti Tak langsung
- 4. (𝑴 𝒔) ∙ (𝑺 𝑩) ∙ (𝑹 𝑨,𝝋 𝟏 ) ∙ (𝑮 𝑷𝑸. 𝑴 𝒕) ∙ (𝑮 𝑲𝑳)(𝚫𝑿𝒀𝒁) Komposisi 5 Isometri Dasar
- 5. 1 2 3 4 5 6 Bukti Langsung 5 (𝑴 𝒔) ∙ (𝑺 𝑩) ∙ (𝑹 𝑨,𝝋 𝟏 ) ∙ (𝑮 𝑷𝑸. 𝑴 𝒕) ∙ (𝑮 𝑲𝑳)(𝚫𝑿𝒀𝒁) = (𝑴 𝒔) ∙ (𝑺 𝑩) ∙ (𝑹 𝑨,𝝓 𝟏 ) ∙ (𝑮 𝑷𝑸. 𝑴 𝒕)(𝚫𝑿 𝟏 𝒀 𝟏 𝒁 𝟏) = 𝑴 𝒔 ∙ (𝑺 𝑩) ∙ (𝑹 𝑨,𝝓 𝟏 )(𝚫𝐗 𝟑 𝐘𝟑 𝐙 𝟑) = (𝑴 𝒔) ∙ (𝑺 𝑩) (𝚫𝑿 𝟒 𝒀 𝟒 𝒁 𝟒) = (𝑴 𝒔)(𝚫𝑿 𝟓 𝒀 𝟓 𝒁 𝟓) = (𝚫𝑿 𝟔 𝒀 𝟔 𝒁 𝟔)
- 6. 𝐿 𝐾 𝑋 𝑌 𝑍 𝑋1 𝑍1 𝑌1 (𝑮 𝑲𝑳)(𝚫𝑿𝒀𝒁) Geseran daerah segitiga XYZ terhadap ruas garis berarah KL
- 7. 𝑡 𝐿 𝐾 𝑍 𝑌 𝑋 𝑋1 𝑍1 𝑌2 𝑍2 𝑌1 𝑋2 (𝑮 𝑷𝑸. 𝑴 𝒕)(𝚫𝑿 𝟏 𝒀 𝟏 𝒁 𝟏) Refleksi Geser daerah segitiga 𝑿 𝟏 𝒀 𝟏 𝒁 𝟏
- 9. 𝑡 𝑄 𝑃 𝐿 𝐾 𝑋3 𝑌3 𝑍3 𝑋1 𝑍1 𝑌2 𝑌1 𝑍2 𝑋2 𝑋 𝑌 𝑍 𝑋4 𝑌4 𝑍4 𝐴 𝜑 𝜑 (𝑹 𝑨,𝝋 𝟏 )(𝚫𝑿 𝟑 𝒀 𝟑 𝒁 𝟑) Rotasi daerah segitiga 𝑿 𝟑 𝒀 𝟑 𝒁 𝟑 terhadap titik A
- 10. P Q 𝐿 𝐾 𝐴 𝑋4 𝑌4 𝑍4 𝑋3 𝑌3 𝑍3 𝑋1 𝑍1 𝑌2 𝑋2 𝑌1 𝑍2 𝑍 𝑋 𝑌 𝐵 𝑡 𝑋5 𝑌5 𝑍5 𝑺 𝑩(𝚫𝑿 𝟒 𝒀 𝟒 𝒁 𝟒) Setengah putaran daerah segitiga 𝑿 𝟒 𝒀 𝟒 𝒁 𝟒 terhadap titik B
- 11. 𝑡 𝑆 𝐿 𝐾 Q P 𝐵 𝐴 𝑌4 𝑋4 𝑍4 𝑋3 𝑌3 𝑍3 𝑋2 𝑍2 𝑌1 𝑌2 𝑋1 𝑍1 𝑋 𝑍 𝑌 𝑋6 𝑍6 𝑌6 𝑌5 𝑋5 𝑍5 𝑴 𝒔(𝚫𝑿 𝟓 𝒀 𝟓 𝒁 𝟓) Pencerminan daerah segitiga 𝑿 𝟓 𝒀 𝟓 𝒁 𝟓 terhadap garis s
- 12. 1 2 3 4 5 6 12 Bukti Tak Langsung 𝑴 𝒔 ∙ 𝑺 𝑩 ∙ 𝑹 𝑨,𝝋 𝟏 ∙ 𝑮 𝑷𝑸. 𝑴 𝒕 ∙ 𝑮 𝑲𝑳 … 𝟏) = 𝑴 𝒔 ∙ 𝑴 𝒈 ∙ 𝑴 𝒇 ∙ 𝑴 𝒇. 𝑴 𝒆 ∙ 𝑴 𝒅 ∙ 𝑴 𝒄 ∙ 𝑴 𝒕 ∙ 𝑴 𝒃 ∙ 𝑴 𝒂 … 𝟐) = (𝑴𝒔 ∙ 𝑴 𝒈) ∙ 𝑰 ∙ (𝑴𝒆 ∙ 𝑴 𝒅) ∙ 𝑴 𝒄 ∙ (𝑴𝒕 ∙ 𝑴𝒃) ∙ 𝑴 𝒂 = 𝑹 𝑬,𝝋 𝟒 ∙ 𝑹 𝑫,𝝋 𝟑 ∙ 𝑴 𝒄 ∙ 𝑹 𝑪,𝝋 𝟐 ∙ 𝑴 𝒂 …3) = (𝑴𝒋 ∙ 𝑴𝒊) ∙ (𝑴𝒊 ∙ 𝑴 𝒉) ∙ 𝑴 𝒄 ∙ ( 𝑹 𝑪,𝝋 𝟐 ) ∙ 𝑴 𝒂 …4) = 𝑴𝒋 ∙ (𝑴 𝒉 ∙ 𝑴 𝒄) ∙ ( 𝑹 𝑪,𝝋 𝟐 ) ∙ 𝑴 𝒂 = 𝑴𝒋 ∙ (𝑹 𝑭,𝝋 𝟓 ) ∙ ( 𝑹 𝑪,𝝋 𝟐 ) ∙ 𝑴 𝒂 …5) = 𝑴𝒋 ∙ (𝑴 𝒎 ∙ 𝑴𝒍) ∙ (𝑴𝒍 ∙ 𝑴 𝒌) ∙ 𝑴 𝒂 …6) = (𝑴𝒋 ∙ 𝑴 𝒎) ∙ (𝑴𝒌 ∙ 𝑴 𝒂) = (𝑹 𝑮,𝝋 𝟔 ) ∙ (𝑹 𝑯,𝝋 𝟕 ) …7) = (𝑴 𝒑 ∙ 𝑴 𝒐) ∙ (𝑴𝒐 ∙ 𝑴 𝒏) …8) = (𝑴 𝒑 ∙ 𝑴 𝒏) = (𝑹 𝑰,𝝋 𝟖 ) …9)
- 13. 1 2 3 4 5 6 13 Syarat Bukti Tak Langsu ng 1) 𝑷𝑸 ∥ 𝒕 2) 𝐚 ∥ 𝒃; 𝒂, 𝒃 ⊥ 𝑲𝑳; 𝒅 𝒂,𝒃 = 𝟏 𝟐 𝑲𝑳; 𝒄 ∥ 𝒅; 𝒄, 𝒅 ⊥ 𝑷𝑸; 𝒅 𝒄,𝒅 = 𝟏 𝟐 𝑷𝑸; 𝑨, 𝑩 ∈ 𝒇; 𝒎∠ 𝒆, 𝒇 = 𝟏 𝟐 𝝋 𝟏; 𝒆 ∩ 𝒇 = 𝑨; 𝒇 ⊥ 𝒈; 𝒇 ∩ 𝒈 = 𝑨 3) 𝒃 ∩ 𝒕 = 𝑪; 𝒎∠ 𝒃, 𝒕 = 𝟏 𝟐 𝝋 𝟐; 𝐝 ∩ 𝒆 = 𝑫; 𝒎∠ 𝒅, 𝒆 = 𝟏 𝟐 𝝋 𝟑; 𝒈 ∩ 𝒔 = 𝐄; 𝒎∠ 𝒈, 𝒔 = 𝟏 𝟐 𝝋 𝟒 4) 𝑫, 𝑬 ∈ 𝒊; 𝒋 ∩ 𝒊 = 𝐄; 𝒎∠ 𝒊, 𝒋 = 𝟏 𝟐 𝝋 𝟒; 𝒉 ∩ 𝒊 = 𝑫; 𝒎∠ 𝒉, 𝒊 = 𝟏 𝟐 𝝋 𝟑 5) 𝒄 ∩ 𝒉 = 𝑭; 𝒎∠ 𝒄, 𝒉 = 𝟏 𝟐 𝝋 𝟓 6) 𝑭, 𝑪 ∈ 𝒍; 𝒍 ∩ 𝒎 = 𝑭; 𝒎∠ 𝒍, 𝒎 = 𝟏 𝟐 𝝋 𝟓; 𝒌 ∩ 𝒍 = 𝑪; 𝒎∠ 𝒌, 𝒍 = 𝟏 𝟐 𝝋 𝟐 7) 𝒎 ∩ 𝒋 = 𝑮; 𝒎∠ 𝒎, 𝒋 = 𝟏 𝟐 𝝋 𝟔; 𝒂 ∩ 𝒌 = 𝑯; 𝒎∠ 𝒂, 𝒌 = 𝟏 𝟐 𝝋 𝟕 8) 𝑮, 𝑯 ∈ 𝒐; 𝒐 ∩ 𝒑 = 𝑮; 𝒎∠ 𝒐, 𝒑 = 𝟏 𝟐 𝝋 𝟔; 𝒏 ∩ 𝒐 = 𝑯; 𝒎∠ 𝒏, 𝒐 = 𝟏 𝟐 𝝋 𝟕 9) 𝒏 ∩ 𝒑 = 𝑰; 𝒎∠ 𝒏, 𝒑 = 𝟏 𝟐 𝝋 𝟖