Diketahui segi tiga ABC, dengan panjang sisi AC = 12 cm, sudut ACB = 75°, sudut ABC = 45°. Tentukan keliling dan luas segi tiga ABC.

Diketahui segi tiga ABC, dengan panjang sisi AC = 12 cm, sudut ACB = 75°, sudut ABC = 45°. Tentukan keliling dan luas segi tiga ABC.

JAwaban : K = 12 + 6(√2 + √3 + √6) cm dan L = (54 + 18√6) cm².

Perhatikan penjelasan berikut ya.
Ingat :
1) Aturan sinus pada perbandingan sisi segitiga AB yaitu :
AB/sin ∠ACB = BC/sin ∠BAC = AC/sin ∠ABC
2) Rumus keliling segitiga ABC yaitu K = AB + BC + AC.
3) Rumus luas segitiga yaitu L = 1/2 × AB × BC × sin∠ABC.

Diketahui segitiga ABC dengan :
AC = 12 cm.
ACB = 75°.
ABC = 45°.
Tentukan keliling dan luas segitiga.

Tentukan panjang sisi AB.
AB/sin ∠ACB = AC/sin ∠ABC
AB/sin 75° = 12/sin 45°
AB(sin 45°) = 12(sin 75°)
AB(√2/2) = 12 · sin(45° + 30°)
AB(√2/2) = 12 · (sin 45°·cos 30° + cos 45°·sin 30°)
AB(√2/2) = 12 ·(√2/2 · √3/2 + √2/2 + 1/2)
AB(√2/2) = 12 · (√6/4 + √2/4)
AB(√2/2) = 12(√6 + √4)/4
AB(√2/2) = 3(√6 + √4)
AB(√2) = 6(√6 + √4)
AB(√2) = 6√6 + 6√4
AB = (6√6 + 6√4)/√2
AB = (6√3 + 6√2) cm

Sebelumnya tentukan besar sudut BAC, untuk menentukan panjang sisi BC.
∠BAC = 180° – ∠ACB – ∠ABC
∠BAC = 180° – 75° – 45°
∠BAC = 60°

BC/sin ∠BAC = AC/sin ∠ABC
BC/sin 60° = 12/sin 45°
BC(sin 45°) = 12(sin 60°)
BC(√2/2) = 12(√3/2)
BC(√2/2) = 6√3
BC(√2) = 2(6√3)
BC = 12√3/√2 → rasionalkan kali √2/√2
BC = (12√3 × √2)/(√2 × √2)
BC = 12√6/2
BC = 6√6 cm

Tentukan keliling segitiga.
K = AB + BC + AC
K = (6√3 + 6√2) + 6√6 + 12
K = 12 + 6(√2 + √3 + √6) cm

Tentukan luas segitiga.
L = 1/2 × AB × BC × sin∠ABC
L = 1/2 × (6√3 + 6√2) × 6√6 × sin 45°
L = (3√3 + 3√2) × 6√6 × √2/2
L = (3√3 + 3√2) × 3√12
L = (3√3 + 3√2) × 3(2√3)
L = (3√3 + 3√2) × 6√3
L = 18(3) + 18√6
L = (54 + 18√6) cm²

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Jadi, keliling dan luas segitiga berturut – turut adalah K = 12 + 6(√2 + √3 + √6) cm dan L = (54 + 18√6) cm².

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