tolong ya terimakasih!
A. sudut pusat = 45⁰, r = 42 cm
K = 2π r
[tex] = 2. \frac{22}{7} .42 \\ = 264 \: cm[/tex]
[tex] \frac{sduut \: pusat}{360 {}^{0} } = \frac{p.busur}{k.lingkaran} \\ \frac{45 {}^{0} }{360 {}^{0} } = \frac{p.busur}{264} \\ \frac{1}{8} = \frac{p.busur}{264} [/tex]
264 = 8 P.busur
P.busur = 264/8 = 33 cm
B. sudut pusat = 90⁰, L.juring = 1256 cm²
[tex] \frac{s.pusat}{360 {}^{0} } = \frac{l.juring}{l.lingkaran} \\ \frac{90 {}^{0} }{360 {}^{0} } = \frac{1256}{l.lingkaran} \\ \frac{1}{4} = \frac{1256}{l.lingkaran} [/tex]
1256.4 = L.lingkaran
5024 cm² = L.lingkaran
L = π r²
5024 = 3,14 r²
r² = 5024/3,14
r² = 1600
[tex]r = \sqrt{1600} = 40 \: cm[/tex]
