This document summarizes the steps to draw the shadow of a triangle under five isometric transformations. It begins by stating the given information: a triangle ΔUVW and the transformations (S B)(R A,φ1)(G KL)(G PQ. Mt)(MS). It then shows each step of applying the transformations to arrive at the final transformed triangle R M,φ9(ΔUVW). The proof is done indirectly by relating the angles and lines created at each step. Diagrams are provided to illustrate the transformations.

1. TUGAS AKHIR
GEOMETRI TRANSFORMASI

2. Oleh:
1. Ria Risqiana Agustina 4101415015
2. Siti Nurzulifa 4101415030
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)

3. Diketahui:
Ditanya:
Lukislah (𝑆 𝐵)(𝑅 𝐴,𝜑1
)(𝐺 𝐾𝐿)(𝐺 𝑃𝑄. 𝑀𝑡)(𝑀𝑆)(∆𝑈𝑉𝑊)
1. 𝐾𝐿 sembarang,
2. Garis s ⊥ t,
3. 𝑃𝑄 ∕∕ 𝑡
4. 𝑔 ⊥ 𝑠, 𝑔 ∩ 𝑠 = {𝐵} , 𝐵 ∈ 𝑠
5. 𝑒 ∩ 𝑓 = {𝐴}, 𝑚 ∠𝑒, 𝑓 =
1
2
𝜑1

4. 𝜑1
t
𝑒
f
A
s
B
1
2
𝜑1
K
L

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

6. 𝜑1
𝑈
𝑉
𝑊
𝑉5
𝑊5
t
𝑈1
𝑉1
𝑊1
𝑈2
𝑉2
𝑊2
𝑒
f
A𝑊4
𝑈4
𝑉4
𝑊3
𝑉3
𝑈3
s
B
𝑊6
𝑉6
1
2
𝜑1
K
L
𝑈5
𝑈6

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

8. 𝜑1
t
𝑒
f
A
s
B
1
2
𝜑1
K
L
ab

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

11. 𝜑1
𝑈
𝑉
𝑊
𝑉5
𝑊5
t
𝑈1
𝑉1
𝑊1
𝑈2
𝑉2
𝑊2
𝑒
f
A𝑊4
𝑈4
𝑉4
𝑊3
𝑉3
𝑈3
s
B
𝑊6
𝑉6
1
2
𝜑1
K
L
𝑈5
𝑈6
ab

12. t
𝑒
f
A
s
K
L
abl
1
2
𝜑4
1
2
𝜑4
mk
G
B C
D
E
1
2
𝜑6
o
1
2
𝜑7
p
I
H
90°
F
M
𝜑9
𝑈
𝑉
𝑊
𝑊6
𝑉6
𝑈6
1
2
𝜑9
J