Kk.. Bantu saya ya kk..Pakai cara.. Biar hari Senin saya

Berikut ini adalah pertanyaan dari milkyind16 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Kk.. Bantu saya ya kk..Pakai cara..
Biar hari Senin saya paham cara mengerjakannya. Makasih..
Dari no.3 sampai 9 ya kk (kecuali no.5 dan 6)

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

No. 3 dan 4, Gunakan Aturan Sinus

No. 7, 8, dan 9, Gunakan Sifat Pythagoras

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ATURAN SINUS

Lihat Gambar 1

$\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\$

$\\$

PYTHAGORAS

Lihat Gambar 2

BC² = AB² + AC²

AB² = BC² – AC²

AC² = BC² – AB²

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==================================

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3.

Diketahui:

∠B = 60°

∠C = 45°

BD = 5 cm

Ditanya:

a. AB = ?

b. AD = ?

Jawab:

a. Perhatikan segitiga ABD (Gambar 3)

Dari segitiga tersebut, diketahui

BC = 5 cm

∠B = 60°

∠D = 90°

maka, ∠A = 180° − 60° − 90° = 30°

Dengan menggunakan Aturan Sinus, diperoleh:

$\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\$

$\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\$

$\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\$

$AB = 5 \times \frac{2}{1}\\$

$AB = 5 \times 2\\$

$AB = 10 \: \text{cm}\\$

$\\$

b. Perhatikan segitiga ABD (Gambar 3)

Dengan menggunakan Aturan Sinus, diperoleh:

$\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\$

$\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\$

$\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\$

$\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\$

$AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\$

$AD = 5 \times \sqrt{3}\\$

$AD = 5\sqrt{3} \: \text{cm}\\$

$\\$

4.

Diketahui:

∠B = 60°

∠C = 45°

AD = 15 cm

Ditanya:

AC = ?

Jawab:

Perhatikan segitiga ADC (Gambar 4)

AD = 15 cm

∠D = 90°

∠C = 45°

Dengan menggunakan Aturan Sinus, diperoleh:

$\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\$

$\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\$

$\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\$

$AC = \frac{15}{\frac{\sqrt{2}}{2}}\\$

$AC = 15 \times \frac{2}{\sqrt{2}}\\$

$AC = \frac{30}{\sqrt{2}}\\$

$AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\$

$AC = \frac{30\sqrt{2}}{2}\\$

$AC = 15\sqrt{2} \: \text{cm}\\$

$\\$

7.

Diketahui:

Kubus ABCD.EFGH dengan panjang rusuk 200 cm

Artinya

AB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cm

Ditanya:

AC = ?

EC = ?

Jawab:

▪︎Menentukan panjang AC (Diagonal Bidang)

AC² = AB² + AC²

AC² = 200² + 200²

AC² = 40.000 + 40.000

AC² = 80.000

AC = √80.000

AC = √(40.000×2)

AC = √40.000 √2

AC = 200√2 cm

▪︎Menentukan panjang EC

EC² = AC² + AE²

EC² = (200√2)² + 200²

EC² = 80.000 + 40.000

EC² = 120.000

EC = √120.000

EC = √(40.000×3)

EC = √40.000 √3

EC = 200√3 cm

$\\$

8.

Diketahui:

AB = 13 cm

BC = 4 cm

CD = 3 cm

Ditanya:

AD = ?

Jawab:

AD² = AB² - BD²

Tentukan nilai BD terlebih dahulu

BD² = BC² + CD²

BD² = 4² + 3²

BD² = 16 + 9

BD² = 25

BD = √25

BD = 5 cm

maka

AD² = AB² − BD²

AD² = 13² − 5²

AD² = 169 − 25

AD² = 144

AD = √144

AD = 12 cm

$\\$

9.

Diketahui:

AB = 12 cm

BC = 9 cm

CD = 8 cm

Ditanya:

AD = ?

Jawab:

AD² = AC² + CD²

Tentukan nilai AC terlebih dahulu

AC² = AB² + BC²

AC² = 12² + 9²

AC² = 144 + 81

AC² = 225

AC = √225

AC = 15 cm

maka

AD² = AC² + CD²

AD² = 15² + 8²

AD² = 225 + 64

AD² = 289

AD = √289

AD = 17 cm

$\\$ $\\$

Semoga membantu

$\\$

==================================

$\\$

Note:

Sifat Akar

• √a × √a = √(a²) = a

• Misal: c = a × b, maka √c = √(a × b) = √a√b

• $\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\$

$No. 3 dan 4, Gunakan Aturan SinusNo. 7, 8, dan 9, Gunakan Sifat Pythagoras[tex]\\[/tex]ATURAN SINUSLihat Gambar 1[tex]\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\[/tex][tex]\\[/tex]PYTHAGORAS Lihat Gambar 2BC² = AB² + AC²AB² = BC² – AC²AC² = BC² – AB²[tex]\\[/tex]==================================[tex]\\[/tex]3.Diketahui:∠B = 60°∠C = 45°BD = 5 cmDitanya:a. AB = ?b. AD = ?Jawab:a. Perhatikan segitiga ABD (Gambar 3)Dari segitiga tersebut, diketahui BC = 5 cm∠B = 60°∠D = 90°maka, ∠A = 180° − 60° − 90° = 30°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AB = 5 \times \frac{2}{1}\\[/tex][tex]AB = 5 \times 2\\[/tex][tex]AB = 10 \: \text{cm}\\[/tex][tex]\\[/tex]b. Perhatikan segitiga ABD (Gambar 3)Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\[/tex][tex]AD = 5 \times \sqrt{3}\\[/tex][tex]AD = 5\sqrt{3} \: \text{cm}\\[/tex][tex]\\[/tex]4.Diketahui:∠B = 60°∠C = 45°AD = 15 cmDitanya:AC = ?Jawab:Perhatikan segitiga ADC (Gambar 4)AD = 15 cm∠D = 90°∠C = 45°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\[/tex][tex]\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\[/tex][tex]\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\[/tex][tex]AC = \frac{15}{\frac{\sqrt{2}}{2}}\\[/tex][tex]AC = 15 \times \frac{2}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\[/tex][tex]AC = \frac{30\sqrt{2}}{2}\\[/tex][tex]AC = 15\sqrt{2} \: \text{cm}\\[/tex][tex]\\[/tex]7.Diketahui:Kubus ABCD.EFGH dengan panjang rusuk 200 cmArtinyaAB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cmDitanya:AC = ?EC = ?Jawab:▪︎Menentukan panjang AC (Diagonal Bidang)AC² = AB² + AC²AC² = 200² + 200²AC² = 40.000 + 40.000AC² = 80.000AC = √80.000AC = √(40.000×2)AC = √40.000 √2AC = 200√2 cm▪︎Menentukan panjang ECEC² = AC² + AE²EC² = (200√2)² + 200²EC² = 80.000 + 40.000EC² = 120.000EC = √120.000EC = √(40.000×3)EC = √40.000 √3EC = 200√3 cm[tex]\\[/tex]8.Diketahui: AB = 13 cmBC = 4 cmCD = 3 cmDitanya:AD = ?Jawab:AD² = AB² - BD²Tentukan nilai BD terlebih dahuluBD² = BC² + CD²BD² = 4² + 3²BD² = 16 + 9BD² = 25BD = √25BD = 5 cmmakaAD² = AB² − BD²AD² = 13² − 5²AD² = 169 − 25AD² = 144AD = √144AD = 12 cm[tex]\\[/tex]9. Diketahui:AB = 12 cmBC = 9 cmCD = 8 cmDitanya:AD = ?Jawab: AD² = AC² + CD²Tentukan nilai AC terlebih dahuluAC² = AB² + BC²AC² = 12² + 9²AC² = 144 + 81AC² = 225AC = √225AC = 15 cmmakaAD² = AC² + CD²AD² = 15² + 8²AD² = 225 + 64AD² = 289AD = √289AD = 17 cm[tex]\\[/tex][tex]\\[/tex]Semoga membantu [tex]\\[/tex]==================================[tex]\\[/tex]Note:Sifat Akar• √a × √a = √(a²) = a• Misal: c = a × b, maka √c = √(a × b) = √a√b• [tex]\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\[/tex]$ $No. 3 dan 4, Gunakan Aturan SinusNo. 7, 8, dan 9, Gunakan Sifat Pythagoras[tex]\\[/tex]ATURAN SINUSLihat Gambar 1[tex]\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\[/tex][tex]\\[/tex]PYTHAGORAS Lihat Gambar 2BC² = AB² + AC²AB² = BC² – AC²AC² = BC² – AB²[tex]\\[/tex]==================================[tex]\\[/tex]3.Diketahui:∠B = 60°∠C = 45°BD = 5 cmDitanya:a. AB = ?b. AD = ?Jawab:a. Perhatikan segitiga ABD (Gambar 3)Dari segitiga tersebut, diketahui BC = 5 cm∠B = 60°∠D = 90°maka, ∠A = 180° − 60° − 90° = 30°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AB = 5 \times \frac{2}{1}\\[/tex][tex]AB = 5 \times 2\\[/tex][tex]AB = 10 \: \text{cm}\\[/tex][tex]\\[/tex]b. Perhatikan segitiga ABD (Gambar 3)Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\[/tex][tex]AD = 5 \times \sqrt{3}\\[/tex][tex]AD = 5\sqrt{3} \: \text{cm}\\[/tex][tex]\\[/tex]4.Diketahui:∠B = 60°∠C = 45°AD = 15 cmDitanya:AC = ?Jawab:Perhatikan segitiga ADC (Gambar 4)AD = 15 cm∠D = 90°∠C = 45°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\[/tex][tex]\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\[/tex][tex]\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\[/tex][tex]AC = \frac{15}{\frac{\sqrt{2}}{2}}\\[/tex][tex]AC = 15 \times \frac{2}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\[/tex][tex]AC = \frac{30\sqrt{2}}{2}\\[/tex][tex]AC = 15\sqrt{2} \: \text{cm}\\[/tex][tex]\\[/tex]7.Diketahui:Kubus ABCD.EFGH dengan panjang rusuk 200 cmArtinyaAB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cmDitanya:AC = ?EC = ?Jawab:▪︎Menentukan panjang AC (Diagonal Bidang)AC² = AB² + AC²AC² = 200² + 200²AC² = 40.000 + 40.000AC² = 80.000AC = √80.000AC = √(40.000×2)AC = √40.000 √2AC = 200√2 cm▪︎Menentukan panjang ECEC² = AC² + AE²EC² = (200√2)² + 200²EC² = 80.000 + 40.000EC² = 120.000EC = √120.000EC = √(40.000×3)EC = √40.000 √3EC = 200√3 cm[tex]\\[/tex]8.Diketahui: AB = 13 cmBC = 4 cmCD = 3 cmDitanya:AD = ?Jawab:AD² = AB² - BD²Tentukan nilai BD terlebih dahuluBD² = BC² + CD²BD² = 4² + 3²BD² = 16 + 9BD² = 25BD = √25BD = 5 cmmakaAD² = AB² − BD²AD² = 13² − 5²AD² = 169 − 25AD² = 144AD = √144AD = 12 cm[tex]\\[/tex]9. Diketahui:AB = 12 cmBC = 9 cmCD = 8 cmDitanya:AD = ?Jawab: AD² = AC² + CD²Tentukan nilai AC terlebih dahuluAC² = AB² + BC²AC² = 12² + 9²AC² = 144 + 81AC² = 225AC = √225AC = 15 cmmakaAD² = AC² + CD²AD² = 15² + 8²AD² = 225 + 64AD² = 289AD = √289AD = 17 cm[tex]\\[/tex][tex]\\[/tex]Semoga membantu [tex]\\[/tex]==================================[tex]\\[/tex]Note:Sifat Akar• √a × √a = √(a²) = a• Misal: c = a × b, maka √c = √(a × b) = √a√b• [tex]\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\[/tex]$ $No. 3 dan 4, Gunakan Aturan SinusNo. 7, 8, dan 9, Gunakan Sifat Pythagoras[tex]\\[/tex]ATURAN SINUSLihat Gambar 1[tex]\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\[/tex][tex]\\[/tex]PYTHAGORAS Lihat Gambar 2BC² = AB² + AC²AB² = BC² – AC²AC² = BC² – AB²[tex]\\[/tex]==================================[tex]\\[/tex]3.Diketahui:∠B = 60°∠C = 45°BD = 5 cmDitanya:a. AB = ?b. AD = ?Jawab:a. Perhatikan segitiga ABD (Gambar 3)Dari segitiga tersebut, diketahui BC = 5 cm∠B = 60°∠D = 90°maka, ∠A = 180° − 60° − 90° = 30°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AB = 5 \times \frac{2}{1}\\[/tex][tex]AB = 5 \times 2\\[/tex][tex]AB = 10 \: \text{cm}\\[/tex][tex]\\[/tex]b. Perhatikan segitiga ABD (Gambar 3)Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\[/tex][tex]AD = 5 \times \sqrt{3}\\[/tex][tex]AD = 5\sqrt{3} \: \text{cm}\\[/tex][tex]\\[/tex]4.Diketahui:∠B = 60°∠C = 45°AD = 15 cmDitanya:AC = ?Jawab:Perhatikan segitiga ADC (Gambar 4)AD = 15 cm∠D = 90°∠C = 45°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\[/tex][tex]\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\[/tex][tex]\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\[/tex][tex]AC = \frac{15}{\frac{\sqrt{2}}{2}}\\[/tex][tex]AC = 15 \times \frac{2}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\[/tex][tex]AC = \frac{30\sqrt{2}}{2}\\[/tex][tex]AC = 15\sqrt{2} \: \text{cm}\\[/tex][tex]\\[/tex]7.Diketahui:Kubus ABCD.EFGH dengan panjang rusuk 200 cmArtinyaAB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cmDitanya:AC = ?EC = ?Jawab:▪︎Menentukan panjang AC (Diagonal Bidang)AC² = AB² + AC²AC² = 200² + 200²AC² = 40.000 + 40.000AC² = 80.000AC = √80.000AC = √(40.000×2)AC = √40.000 √2AC = 200√2 cm▪︎Menentukan panjang ECEC² = AC² + AE²EC² = (200√2)² + 200²EC² = 80.000 + 40.000EC² = 120.000EC = √120.000EC = √(40.000×3)EC = √40.000 √3EC = 200√3 cm[tex]\\[/tex]8.Diketahui: AB = 13 cmBC = 4 cmCD = 3 cmDitanya:AD = ?Jawab:AD² = AB² - BD²Tentukan nilai BD terlebih dahuluBD² = BC² + CD²BD² = 4² + 3²BD² = 16 + 9BD² = 25BD = √25BD = 5 cmmakaAD² = AB² − BD²AD² = 13² − 5²AD² = 169 − 25AD² = 144AD = √144AD = 12 cm[tex]\\[/tex]9. Diketahui:AB = 12 cmBC = 9 cmCD = 8 cmDitanya:AD = ?Jawab: AD² = AC² + CD²Tentukan nilai AC terlebih dahuluAC² = AB² + BC²AC² = 12² + 9²AC² = 144 + 81AC² = 225AC = √225AC = 15 cmmakaAD² = AC² + CD²AD² = 15² + 8²AD² = 225 + 64AD² = 289AD = √289AD = 17 cm[tex]\\[/tex][tex]\\[/tex]Semoga membantu [tex]\\[/tex]==================================[tex]\\[/tex]Note:Sifat Akar• √a × √a = √(a²) = a• Misal: c = a × b, maka √c = √(a × b) = √a√b• [tex]\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\[/tex]$ $No. 3 dan 4, Gunakan Aturan SinusNo. 7, 8, dan 9, Gunakan Sifat Pythagoras[tex]\\[/tex]ATURAN SINUSLihat Gambar 1[tex]\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\[/tex][tex]\\[/tex]PYTHAGORAS Lihat Gambar 2BC² = AB² + AC²AB² = BC² – AC²AC² = BC² – AB²[tex]\\[/tex]==================================[tex]\\[/tex]3.Diketahui:∠B = 60°∠C = 45°BD = 5 cmDitanya:a. AB = ?b. AD = ?Jawab:a. Perhatikan segitiga ABD (Gambar 3)Dari segitiga tersebut, diketahui BC = 5 cm∠B = 60°∠D = 90°maka, ∠A = 180° − 60° − 90° = 30°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AB = 5 \times \frac{2}{1}\\[/tex][tex]AB = 5 \times 2\\[/tex][tex]AB = 10 \: \text{cm}\\[/tex][tex]\\[/tex]b. Perhatikan segitiga ABD (Gambar 3)Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\[/tex][tex]AD = 5 \times \sqrt{3}\\[/tex][tex]AD = 5\sqrt{3} \: \text{cm}\\[/tex][tex]\\[/tex]4.Diketahui:∠B = 60°∠C = 45°AD = 15 cmDitanya:AC = ?Jawab:Perhatikan segitiga ADC (Gambar 4)AD = 15 cm∠D = 90°∠C = 45°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\[/tex][tex]\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\[/tex][tex]\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\[/tex][tex]AC = \frac{15}{\frac{\sqrt{2}}{2}}\\[/tex][tex]AC = 15 \times \frac{2}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\[/tex][tex]AC = \frac{30\sqrt{2}}{2}\\[/tex][tex]AC = 15\sqrt{2} \: \text{cm}\\[/tex][tex]\\[/tex]7.Diketahui:Kubus ABCD.EFGH dengan panjang rusuk 200 cmArtinyaAB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cmDitanya:AC = ?EC = ?Jawab:▪︎Menentukan panjang AC (Diagonal Bidang)AC² = AB² + AC²AC² = 200² + 200²AC² = 40.000 + 40.000AC² = 80.000AC = √80.000AC = √(40.000×2)AC = √40.000 √2AC = 200√2 cm▪︎Menentukan panjang ECEC² = AC² + AE²EC² = (200√2)² + 200²EC² = 80.000 + 40.000EC² = 120.000EC = √120.000EC = √(40.000×3)EC = √40.000 √3EC = 200√3 cm[tex]\\[/tex]8.Diketahui: AB = 13 cmBC = 4 cmCD = 3 cmDitanya:AD = ?Jawab:AD² = AB² - BD²Tentukan nilai BD terlebih dahuluBD² = BC² + CD²BD² = 4² + 3²BD² = 16 + 9BD² = 25BD = √25BD = 5 cmmakaAD² = AB² − BD²AD² = 13² − 5²AD² = 169 − 25AD² = 144AD = √144AD = 12 cm[tex]\\[/tex]9. Diketahui:AB = 12 cmBC = 9 cmCD = 8 cmDitanya:AD = ?Jawab: AD² = AC² + CD²Tentukan nilai AC terlebih dahuluAC² = AB² + BC²AC² = 12² + 9²AC² = 144 + 81AC² = 225AC = √225AC = 15 cmmakaAD² = AC² + CD²AD² = 15² + 8²AD² = 225 + 64AD² = 289AD = √289AD = 17 cm[tex]\\[/tex][tex]\\[/tex]Semoga membantu [tex]\\[/tex]==================================[tex]\\[/tex]Note:Sifat Akar• √a × √a = √(a²) = a• Misal: c = a × b, maka √c = √(a × b) = √a√b• [tex]\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\[/tex]$ $No. 3 dan 4, Gunakan Aturan SinusNo. 7, 8, dan 9, Gunakan Sifat Pythagoras[tex]\\[/tex]ATURAN SINUSLihat Gambar 1[tex]\frac{AB}{\sin {C}} = \frac{AC}{\sin {B}} = \frac{BC}{\sin {A}}\\[/tex][tex]\\[/tex]PYTHAGORAS Lihat Gambar 2BC² = AB² + AC²AB² = BC² – AC²AC² = BC² – AB²[tex]\\[/tex]==================================[tex]\\[/tex]3.Diketahui:∠B = 60°∠C = 45°BD = 5 cmDitanya:a. AB = ?b. AD = ?Jawab:a. Perhatikan segitiga ABD (Gambar 3)Dari segitiga tersebut, diketahui BC = 5 cm∠B = 60°∠D = 90°maka, ∠A = 180° − 60° − 90° = 30°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AB}{\sin {D}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AB}{\sin {90\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AB}{1} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AB = 5 \times \frac{2}{1}\\[/tex][tex]AB = 5 \times 2\\[/tex][tex]AB = 10 \: \text{cm}\\[/tex][tex]\\[/tex]b. Perhatikan segitiga ABD (Gambar 3)Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AD}{\sin {B}} = \frac{BD}{\sin {A}}\\[/tex][tex]\frac{AD}{\sin {60\degree}} = \frac{5}{\sin {30\degree}}\\[/tex][tex]\frac{AD}{\frac{1}{2}\sqrt{3}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]\frac{AD}{\frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{2}}\\[/tex][tex]AD = 5 \times \frac{2}{1} \times \frac{\sqrt{3}}{2}\\[/tex][tex]AD = 5 \times \sqrt{3}\\[/tex][tex]AD = 5\sqrt{3} \: \text{cm}\\[/tex][tex]\\[/tex]4.Diketahui:∠B = 60°∠C = 45°AD = 15 cmDitanya:AC = ?Jawab:Perhatikan segitiga ADC (Gambar 4)AD = 15 cm∠D = 90°∠C = 45°Dengan menggunakan Aturan Sinus, diperoleh:[tex]\frac{AC}{\sin {D}} = \frac{AD}{\sin {C}}\\[/tex][tex]\frac{AC}{\sin {90\degree} = \frac{15}{\sin {45\degree}}\\[/tex][tex]\frac{AC}{1} = \frac{15}{\frac{1}{2} \sqrt{2}}\\[/tex][tex]AC = \frac{15}{\frac{\sqrt{2}}{2}}\\[/tex][tex]AC = 15 \times \frac{2}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}}\\[/tex][tex]AC = \frac{30}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\[/tex][tex]AC = \frac{30\sqrt{2}}{2}\\[/tex][tex]AC = 15\sqrt{2} \: \text{cm}\\[/tex][tex]\\[/tex]7.Diketahui:Kubus ABCD.EFGH dengan panjang rusuk 200 cmArtinyaAB = BC = CD = AD = EF = FG = GH = EH = AE = BF = CG = DH = 200 cmDitanya:AC = ?EC = ?Jawab:▪︎Menentukan panjang AC (Diagonal Bidang)AC² = AB² + AC²AC² = 200² + 200²AC² = 40.000 + 40.000AC² = 80.000AC = √80.000AC = √(40.000×2)AC = √40.000 √2AC = 200√2 cm▪︎Menentukan panjang ECEC² = AC² + AE²EC² = (200√2)² + 200²EC² = 80.000 + 40.000EC² = 120.000EC = √120.000EC = √(40.000×3)EC = √40.000 √3EC = 200√3 cm[tex]\\[/tex]8.Diketahui: AB = 13 cmBC = 4 cmCD = 3 cmDitanya:AD = ?Jawab:AD² = AB² - BD²Tentukan nilai BD terlebih dahuluBD² = BC² + CD²BD² = 4² + 3²BD² = 16 + 9BD² = 25BD = √25BD = 5 cmmakaAD² = AB² − BD²AD² = 13² − 5²AD² = 169 − 25AD² = 144AD = √144AD = 12 cm[tex]\\[/tex]9. Diketahui:AB = 12 cmBC = 9 cmCD = 8 cmDitanya:AD = ?Jawab: AD² = AC² + CD²Tentukan nilai AC terlebih dahuluAC² = AB² + BC²AC² = 12² + 9²AC² = 144 + 81AC² = 225AC = √225AC = 15 cmmakaAD² = AC² + CD²AD² = 15² + 8²AD² = 225 + 64AD² = 289AD = √289AD = 17 cm[tex]\\[/tex][tex]\\[/tex]Semoga membantu [tex]\\[/tex]==================================[tex]\\[/tex]Note:Sifat Akar• √a × √a = √(a²) = a• Misal: c = a × b, maka √c = √(a × b) = √a√b• [tex]\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a}{b}\sqrt{b}\\[/tex]$