tolong bantu ya.terimakasih

MAPEL = MATEMATIKA
KELAS = 8
BAB = GARIS SINGGUNG LINGKARAN

diketahui =
GSPD = 24cm
r1 = 12cm
r2 = 6cm

ditanya = jarak titik pusat (p) ??

[tex]gspd = \sqrt{ {p}^{2} - {(r1 + r2)}^{2} } \\ \: \: \: \: 24 = \sqrt{ {p}^{2} - {(12 + 6)}^{2} } \\ \: \: {24}^{2} = {p}^{2} - {(12 + 6)}^{2} \\ \: 576 = {p}^{2} - 324 \\ 576 + 324 = {p}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: 900 = {p}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p = \sqrt{900} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p = 30[/tex]

jadi jarak titik pusat = 30cm

semoga membantu

Mapel : Matematika
Bab : lingkaran

[tex]gspd = \sqrt{d ^{2} - (r. + r) ^{2} } [/tex]
[tex]24 = \sqrt{d ^{2} - (12 + 6) ^{2} } [/tex]
[tex]24 = \sqrt{d ^{2} - 18^{2} } [/tex]
[tex]24 ^{2} = d ^{2} - 18 ^{2} [/tex]
[tex]576 = d ^{2} - 324[/tex]
[tex]576 + 324 = d ^{2} [/tex]

[tex]900 = d ^{2} [/tex]
[tex] \sqrt{900} = d[/tex]
[tex]30 = d[/tex]






[tex]d = 30 \: cm[/tex]