diketahui perbandingan panjang lebar dan tinggi suatu balok 6:4:5 jika volume balok 7.680cm^3 tentukan p,l,t dan luas permukaan balok?

p : l : t = 6 : 4 : 5
volume = p×l×t
7680 = 6x × 4x × 5x
7680 = 120 x^3
x^3 = 64
x = 4

p= 6×4 = 24
l= 4×4 = 16
t= 5×4 = 20

luas permukaan=2(pl + pt + lt)
luas permukaan= 2(24×16 + 24×20 + 16×20)
luas permukaan= 2(384 + 480 + 320)
luas permukaan = 2368 m kuadrat

jadi luas permukaannya 2368 m kuadrat

maaf kalau salah

[tex]p \div l \div t = 6 \div 4 \div 5[/tex]
maka
[tex]p \div l = 6 \div 4[/tex]
[tex] \: \frac{p}{l} = \frac{6}{4} \\ 4 p = 6l \\ \: \: \: p = \frac{6l}{4} [/tex]
dan

[tex] \: l \div t = 4 \div 5 \\ \\ \frac{l} {t} = \frac{4}{5} \\ 4t = 5l \\ \: \: t = \frac{5l}{4} [/tex]
sehingga

[tex]v = p \times l \times t \\ [/tex]
[tex]7.680 = \frac{6l}{4} \times l \times \frac{5l}{4} \\ 7.680 = \frac{ {30l}^{3} }{16} \\ \: \: \: {30l}^{3} = 7.680 \times 16 \\ \: \: \: {30l}^{3} = 122.880 \\ \: \: \: \: \: \: \: {l}^{3} = \frac{122.880}{30} \\ \: \: \: \: \: \: \: {l}^{3} = 4.096 \\ \: \: \: \: \: \: \: \: {l}^{} = \sqrt[3]{4.096} \\ \: \: \: \: \: \: \: \: l = 16 \: cm[/tex]
[tex]p = \frac{6l}{4} \\ p = \frac{6 \times 16}{4} \\ p = \frac{96}{4} \\ p = 24 \: cm[/tex]
[tex]t = \frac{5l}{4} \\ t = \frac{5 \times 16}{4} \\ t = \frac{80}{4} \\ t = 20 \: cm[/tex]
jadi
luas balok adalah
[tex]2(p \times l + p \times t + l \times t) \\ 2(24 \times 16 + 24 \times 20 + 16 \times 20) \\ 2(384 + 480 + 320) \\ 2(1184) \\ 2368 \: {cm}^{2} [/tex]