Diketahui titik koordinat A(-4,2),B(2,-2),C(8,2),D(2,6) jika ABCD membentuk belah ketupat tentukan keliling belah ketupat?beserta cara?

SUMBU X
AC = C - A
AC = (8,2) - (-4,2)
AC = (8 - (-4),2 - 2)
AC = (8+4,0)
AC = 12 satuan

SUMBU Y
BD = D - B
BD = (2,6) - (2,-2)
BD = (2-2,6-(-2))
BD = (0,6+2)
BD = (0,8)
BD = 8 satuan

AB = BC = CD = DA

AB^2 = ((1/2)AC)^2 + ((1/2)BD)^2
AB^2 = ((1/2)×12)^2 + ((1/2)×8)^2
AB^2 = 6^2 + 4^2
AB^2 = 36 + 16
AB^2 = 52
AB = V52
AB = 7,21

K = 4 × AB
K = 4 × 7,21
K = 28,84 satuan