..
[tex]\begin{gathered}\boxed{ \begin{array}{lr} \boxed{\large{\sf{J= θ × π × r × r}}} \\ \boxed{\large{\sf{B =θ× π × d~~~~~}}}\\ \\\sf{Keterangan \ \: :}\\ \begin{aligned} \sf{ J} &= \sf{ Luas~Juring}\\ \sf{ B } &= \sf{Panjang~Busur} \\ \sf{θ} &= \sf{Besar~Sudut~Pusat} \\\sf{ r } &= \sf{ Jari-Jari }\\ \sf{ d } & =\sf{ Diameter } \end{aligned} \end{array}} \begin{aligned}&~~\to \sf{r = ½d}\\ &~~\to \sf{π=\frac{22}{7} \: jika \: r ~atau~d = kelipatan \: 7} \\&~~\to \sf{π=3.14 \: jika \: r ~atau~d≠ kelipatan \: 7}\end{aligned}\end{gathered}[/tex]
[tex]\begin{aligned} Busur~AB &= \frac{ \angle AOB}{\angle COD} \times busur~CD \\&= \frac{40^{\circ}}{100^{\circ}} \times 40~cm \\&= 0,4 \times 40~cm \\&= \boxed{\bold{\underline{16~cm}}} \end{aligned}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 22 - 05 - 2023}}[/tex]
PB.AB :
= (40/100)(40)
= (40 . 40)/100
= 1.600/100
= 16/1
= 16 CM
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Lingkaran
[Busur]
..
[tex]\begin{gathered}\boxed{ \begin{array}{lr} \boxed{\large{\sf{J= θ × π × r × r}}} \\ \boxed{\large{\sf{B =θ× π × d~~~~~}}}\\ \\\sf{Keterangan \ \: :}\\ \begin{aligned} \sf{ J} &= \sf{ Luas~Juring}\\ \sf{ B } &= \sf{Panjang~Busur} \\ \sf{θ} &= \sf{Besar~Sudut~Pusat} \\\sf{ r } &= \sf{ Jari-Jari }\\ \sf{ d } & =\sf{ Diameter } \end{aligned} \end{array}} \begin{aligned}&~~\to \sf{r = ½d}\\ &~~\to \sf{π=\frac{22}{7} \: jika \: r ~atau~d = kelipatan \: 7} \\&~~\to \sf{π=3.14 \: jika \: r ~atau~d≠ kelipatan \: 7}\end{aligned}\end{gathered}[/tex]
Penyelesaian Soal
[tex]\begin{aligned} Busur~AB &= \frac{ \angle AOB}{\angle COD} \times busur~CD \\&= \frac{40^{\circ}}{100^{\circ}} \times 40~cm \\&= 0,4 \times 40~cm \\&= \boxed{\bold{\underline{16~cm}}} \end{aligned}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 22 - 05 - 2023}}[/tex]
PB.AB :
= (40/100)(40)
= (40 . 40)/100
= 1.600/100
= 16/1
= 16 CM