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[tex]\boxed{ \begin{array}{lr} \boxed{\large{\sf{ LP= 4 \times π \times r²~} }} \\ \boxed{\large{\sf{~V ~ = \frac{4}{3} \times π \times r³ } }}\\ \boxed{\large{\sf{LT = 2 π \times r \times t \: }}} \\ \\ \sf{Keterangan \ \: :}\\ \begin{aligned} \sf{ LP} &= \sf{ Luas~Permukaan }\\ \sf{LT} &= \sf{Luas~Tembereng} \\ \sf{V} &= \sf{ Volume~Bola}\\ \sf{ π } &= \sf{Konstanta~Lingkaran }\\\sf{ r} &= \sf{ Jari-Jari~Bola} \\ \sf{t} &= \sf{Tinggi~Tembereng} \end{aligned} \end{array}}~~\begin{aligned}& \to~r=½d \\ &\to~π=\frac{22}{7}~Jika~r= Kelipatan~7 \\ &\to~π=3.14~Jika~r \ne Kelipatan~7 \end{aligned}[/tex]
[tex]\begin{aligned} LP~Globe &= LP~Bola \\&= 4 \times \pi \times r^2 \\&= 4 \times 3,14 \times (20~cm)^2 \\&= 12,56 \times 400~cm^2 \\&= \boxed{\bold{\underline{5.024,00~cm^2}}} \end{aligned}[/tex]
Opsi Jawaban : B
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 22 - 05 - 2023}}[/tex]
LPB :
= 4πR²
= 4 . 3,14 . (20)²
= 12,56 . (20 . 20)
= 12,56 . 400
= 5.024 CM²
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Bangun Ruang Sisi Lengkung
[Bola]
..
[tex]\boxed{ \begin{array}{lr} \boxed{\large{\sf{ LP= 4 \times π \times r²~} }} \\ \boxed{\large{\sf{~V ~ = \frac{4}{3} \times π \times r³ } }}\\ \boxed{\large{\sf{LT = 2 π \times r \times t \: }}} \\ \\ \sf{Keterangan \ \: :}\\ \begin{aligned} \sf{ LP} &= \sf{ Luas~Permukaan }\\ \sf{LT} &= \sf{Luas~Tembereng} \\ \sf{V} &= \sf{ Volume~Bola}\\ \sf{ π } &= \sf{Konstanta~Lingkaran }\\\sf{ r} &= \sf{ Jari-Jari~Bola} \\ \sf{t} &= \sf{Tinggi~Tembereng} \end{aligned} \end{array}}~~\begin{aligned}& \to~r=½d \\ &\to~π=\frac{22}{7}~Jika~r= Kelipatan~7 \\ &\to~π=3.14~Jika~r \ne Kelipatan~7 \end{aligned}[/tex]
Penyelesaian Soal
[tex]\begin{aligned} LP~Globe &= LP~Bola \\&= 4 \times \pi \times r^2 \\&= 4 \times 3,14 \times (20~cm)^2 \\&= 12,56 \times 400~cm^2 \\&= \boxed{\bold{\underline{5.024,00~cm^2}}} \end{aligned}[/tex]
Opsi Jawaban : B
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 22 - 05 - 2023}}[/tex]
LPB :
= 4πR²
= 4 . 3,14 . (20)²
= 12,56 . (20 . 20)
= 12,56 . 400
= 5.024 CM²