PEMBAHASAN
SOAL NO 6
Menentukan nilai x
[tex]L = \frac{a \times t}{2} [/tex]
[tex]54 \: {cm}^{2} = \frac{12 \: cm \times (x + 3) \: cm}{2} [/tex]
[tex]54 \times 2 = 12 \times (x + 3)[/tex]
[tex]108 = 12x + 36[/tex]
[tex]108 - 36 = 12x[/tex]
[tex]72 \: cm = 12x[/tex]
[tex] \frac{72}{12} \: cm = x[/tex]
[tex]6 \: cm = x[/tex]
menentukan sisi tinggi
t = x + 3 cm
t = 6 cm + 3 cm
t = 9 cm
menentukan sisi miring
[tex]BC = \sqrt{ {9}^{2} + {12}^{2} } [/tex]
[tex]BC = \sqrt{81 + 144} [/tex]
[tex]BC = \sqrt{225} [/tex]
[tex]BC = 15 \: cm[/tex]
menentukan keliling
K = s + s + s
K = 9 cm + 12 cm + 15 cm
K = 36 cm
Kesimpulan:
Jadi, Keliling ∆ABC adalah 36 cm (opsi A).
SOAL NO 7
K = s + s + s
K = 3s
48 cm = 3s
48 cm ÷ 3 = s
16 cm = s
16 cm = a
Menentukan tinggi
[tex]t² = s² - ( \frac{1}{2} \times s)²[/tex]
[tex]t² = 16² - ( \frac{1}{2} \times 16) ²[/tex]
[tex]t² = {16}^{2} - {8}^{2} [/tex]
[tex]t² = 256 - 64[/tex]
[tex]t² = 192[/tex]
[tex]t = \sqrt{192} [/tex]
[tex]t = \sqrt{64 \times 3} [/tex]
[tex]t = \sqrt{ {8}^{2} \times 3} [/tex]
[tex]t = 8 \sqrt{3} \: cm[/tex]
Menentukan Luas segitiga KLM
[tex]L = \frac{a \times t}{2} [/tex]
[tex]L = \frac{16 \: cm \times 8 \sqrt{3} \: cm}{2} [/tex]
[tex]L = 8 \times 8 \sqrt{3} [/tex]
[tex]L = 64 \sqrt{3} \: {cm}^{2} [/tex]
Kesimpulan:
Jadi, Luas ∆KLM adalah 64√3 cm² (opsi D).
No. 6
➢ Nilai x
(x + 3) × 12 cm = 2 × 54 cm²
12x + 36 cm = 108 cm²
12x = 108 - 36
12x = 72
x = 6 cm
➢ Tinggi
= x + 3 cm
= 6 + 3 cm
= 9 cm
➢ Hipotenusa
[tex] \tt = \sqrt{ {9}^{2} + {12}^{2} } [/tex]
[tex] \tt = \sqrt{81 + 144} [/tex]
[tex] \tt = \sqrt{225} [/tex]
= 15 cm
➢ Keliling
= 9 + 12 + 15 cm
= 21 + 15 cm
= 36 cm (A.)
No. 7
- ½ Keliling
= 48 cm ÷ 2
= 24 cm
- Panjang sisi
= 48 cm ÷ 3
= 16 cm
- Luas
[tex] \tt = \sqrt{s(s - a)(s - b)(s - c)} [/tex]
[tex] \tt = \sqrt{24(24 - 16)(24 - 16)(24 - 16)} [/tex]
[tex] \tt = \sqrt{24(8)(8)(8)} [/tex]
[tex] \tt = \sqrt{12.288} [/tex]
= 64√3 cm² (D.)
[tex] \pink{\boxed{\red{\boxed{\purple{\mathfrak{\ast ~ \blue{Celia ~ Claire} ~ \ast}}}}}}[/tex]
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