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1. Ria Risqiana Agustina 4101415015
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3. Dea Amara P. 4101415053
4. Muchamad Idris 4101415091
5. Luluk Syarifatun Ni’mah 4101415132
Melukis Bayangan Bidang Segitiga oleh
Komposisi Lima Isometri
(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. 𝑀𝑡)(𝑀𝑆)
5. (𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. 𝑀𝑡)(𝑀𝑆)(∆𝑈𝑉𝑊)
=(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. 𝑀𝑡)(∆𝑈1 𝑉1 𝑊1)
=(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. [∆𝑈2 𝑉2 𝑊2])
=(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(∆𝑈3 𝑉3 𝑊3)
=(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(∆𝑈4 𝑉4 𝑊4)
=(𝑆 𝐵)(∆𝑈5 𝑉5 𝑊5)
= (∆𝑈6 𝑉6 𝑊6).
Bukti langsung
7. Diketahui :
Akan ditunjukkan bahwa :
(𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. 𝑀𝑡)(𝑀𝑆)(∆𝑈𝑉𝑊) = 𝑅 𝑀,𝜑9
(∆𝑈𝑉𝑊)
1. a//𝑏, 𝑐 ⊥ 𝑃𝑄, 𝑡//𝑃𝑄, 𝑑 𝑎,𝑏 =
1
2
𝑃𝑄
2. 𝑡 ⊥ 𝑠
3. c// 𝑑, 𝑐 ⊥ 𝐾𝐿, 𝑑 𝑐,𝑑 =
1
2
𝐾𝐿
4. 𝑔 ⊥ 𝑠, 𝑔 ∩ 𝑠 = {𝐵} , 𝐵 ∈ 𝑠
5. 𝑒 ∩ 𝑓 = {𝐴}, 𝑚 ∠𝑒, 𝑓 =
1
2
𝜑1
9. Bukti Tak Langsung
𝑆𝐴 ∘ 𝑅 𝐴,𝜑1
∘ 𝐺 𝐾𝐿 ∘ 𝐺 𝑃𝑄 ∘ 𝑀𝑡 ∘ 𝑀𝑠(∆𝑈𝑉𝑊)
= 𝑀𝑆 ∘ 𝑀𝑔 ∘ 𝑀𝑓 ∘ 𝑀𝑒 ∘ 𝑀 𝑑 ∘ 𝑀𝑐 ∘ 𝑀 𝑏 ∘ 𝑀 𝑎 ∘ 𝑀𝑡 ∘ 𝑀𝑠……....(A)
= 𝑀𝑆 ∘ 𝑀𝑔 ∘ 𝑀𝑓 ∘ 𝑀𝑒 ∘ 𝑀 𝑑 ∘ 𝑀𝑐 ∘ 𝑀 𝑏 ∘ 𝑀 𝑎 ∘ 𝑀𝑡 ∘ 𝑀𝑠 .……....(B)
= 𝑀𝑆 ∘ 𝑅 𝐶,𝜑2
∘ 𝑅 𝐷,𝜑3
∘ 𝑅 𝐸,𝜑4
∘ 𝑀 𝑎 ∘ 𝑆 𝐹 …………..…………………....(B)
= 𝑀𝑆 ∘ 𝑀ℎ ∘ 𝑀𝑖 ∘ 𝑀𝑖 ∘ 𝑀𝑗 ∘ 𝑅 𝐸,𝜑4
∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ……………………..(C)
= 𝑀𝑆 ∘ 𝑀ℎ ∘ 𝐼 ∘ 𝑀𝑗 ∘ 𝑅 𝐸,𝜑4
∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ………………………………....(D)
= 𝑀𝑆 ∘ 𝑀ℎ ∘ 𝑀𝑗 ∘ 𝑅 𝐸,𝜑4
∘ 𝑀 𝑎 ∘ 𝑆 𝐹 …………………………….…....(D)
= 𝑀𝑆 ∘ 𝑅 𝐺,𝜑5
∘ 𝑅 𝐸,𝜑4
∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ………………………………………....(D)
= 𝑀𝑆 ∘ 𝑀ℎ ∘ 𝑀𝑗 ∘ 𝑀𝑐 ∘ 𝑀 𝑏 ∘ 𝑀 𝑎 ∘ 𝑆 𝐹……………………………....(E)
= 𝑀𝑆 ∘ 𝑀 𝑘 ∘ 𝑀𝑙 ∘ 𝑀𝑙 ∘ 𝑀 𝑚 ∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ……………………………....(E)
= 𝑀𝑠 ∘ 𝑀 𝑘 ∘ 𝐼 ∘ 𝑀 𝑚 ∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ………………………..…………………………....(F)
= 𝑀𝑠 ∘ 𝑀 𝑘 ∘ 𝑀 𝑚 ∘ 𝑀 𝑎 ∘ 𝑆 𝐹 ……………………….…………………………....(F)
= 𝑅 𝐻,𝜑6
∘ 𝑅𝐼,𝜑7
∘ 𝑆 𝐹 ……………………………………..…..……………………….(F)
= 𝑀 𝑛 ∘ 𝑀 𝑜 ∘ 𝑀 𝑜 ∘ 𝑀 𝑝 ∘ 𝑆 𝐹 ………………………..…………………………....(G)
= 𝑀 𝑛 ∘ 𝐼 ∘ 𝑀 𝑝 ∘ 𝑆 𝐹 ……………………………………..…..…………..……..…....(H)
= 𝑀 𝑛 ∘ 𝑀 𝑝 ∘ 𝑆 𝐹 ……………………………………..…..……………..………....(H)
= 𝑅𝐽,𝜑8
∘ 𝑆 𝐹 ……………………………………….………………………………....(H)
= 𝑀 𝑞 ∘ 𝑀𝑟 ∘ 𝑀𝑟 ∘ 𝑀 𝑢 ……………………………..…..…………………………(I)
= 𝑀 𝑞 ∘ 𝐼 ∘ 𝑀 𝑢 …………………………..………………..…..……………………....(J)
= 𝑀 𝑞 ∘ 𝑀 𝑢 …………………………………………..…..…………………………....(J)
= 𝑅 𝑀,𝜑9
……………………………….……………..…..…………………………....(J)
Jadi, 𝑆𝐴 ∘ 𝑅 𝐴,𝜑1
∘ 𝐺 𝐾𝐿 ∘ 𝐺 𝑃𝑄 ∘ 𝑀𝑡 ∘ 𝑀𝑠 ∆𝑈𝑉𝑊 = 𝑅 𝑀,𝜑9
(∆𝑈𝑉𝑊)
10. A. 1. 𝑠 ⊥ 𝑡
2. 𝑡//𝑃𝑄, 𝑎//𝑏, a ⊥ 𝑃𝑄
3. 𝑑 𝑎,𝑏 =
1
2
𝑃𝑄
4. 𝑐//𝑑, c ⊥ 𝐾𝐿, 𝑑 𝑐,𝑑 =
1
2
𝐾𝐿
5. 𝑒 ∩ 𝑓 = {𝐴}, 𝑚 ∠𝑒, 𝑓 =
1
2
𝜑1
6. g ⊥ 𝑠, 𝑔 ∩ 𝑠 = 𝐵 , 𝐵 ∈ 𝑆
B. 1. 𝑠 ⊥ 𝑡, 𝑠 ∩ 𝑡 = {𝐹}
2. 𝑐 ∩ 𝑏 = {𝐸}, 𝑚 ∠𝑐, 𝑏 =
1
2
𝜑4
3. 𝑑 ∩ 𝑒 = {𝐷}, 𝑚 ∠𝑑, 𝑒 =
1
2
𝜑3
4. 𝑓 ∩ 𝑔 = {𝐶}, 𝑚 ∠𝑓, 𝑔 =
1
2
𝜑2
C. 1. 𝑖 ∩ ℎ = {𝐶}, 𝑚 ∠𝑖, ℎ =
1
2
𝜑2
2. 𝑗 ∩ 𝑖 = {𝐷}, 𝑚 ∠𝑗, 𝑖 =
1
2
𝜑3
D. 1. 𝑗 ∩ ℎ = {𝐺}, 𝑚 ∠𝑗, ℎ =
1
2
𝜑5
E. 1. 𝑗 ∩ ℎ = {𝐺}, 𝑚 ∠𝑗, ℎ =
1
2
𝜑5
2. 𝑏 ∩ 𝑐 = {𝐸}, 𝑚 ∠𝑏, 𝑐 =
1
2
𝜑4
3. 𝑙 ∩ 𝑘 = {𝐺}, 𝑚 ∠𝑙, 𝑘 =
1
2
𝜑5
4. 𝑚 ∩ 𝑙 = {𝐸}, 𝑚 ∠𝑚, 𝑙 =
1
2
𝜑4
F. 1 𝑠 ∩ 𝑘 = {𝐻}, 𝑚 ∠𝑠, 𝑘 =
1
2
𝜑6
2. 𝑎 ∩ 𝑚 = {𝐼}, 𝑚 ∠𝑎, 𝑚 =
1
2
𝜑7
G. 1 𝑜 ∩ 𝑛 = {𝐻}, 𝑚 ∠𝑜, 𝑛 =
1
2
𝜑6
2. 𝑝 ∩ 𝑜 = {𝐼}, 𝑚 ∠𝑝, 𝑜 =
1
2
𝜑7
H. 1 𝑝 ∩ 𝑛 = {𝐽}, 𝑚 ∠𝑝, 𝑛 =
1
2
𝜑8
I. 1 𝑟 ∩ 𝑞 = {𝐽}, 𝑚 ∠𝑟, 𝑞 =
1
2
𝜑8
2. 𝑢 ⊥ 𝑟, 𝑢 ∩ 𝑟 = {𝐹}
J. 1. 𝑢 ∩ 𝑞 = {𝑀}
2. 𝑚 ∠𝑢, 𝑞 =
1
2
𝜑9
Syarat Bukti Tak Langsung