Let PQRS be a square with side length 10. The points X and Y lie outside the square such that XQ = 6, XP = 8, YS = 6, and YR = 8. Find XY^2.
XY^2 = 288.
nice explanation
XY^2 = 392
MX = (8*6)/10 = 4.8
MQ = sqrt(QX² - MX²) = 3.6 MQ = NL
NX = 2*MX + 10 = 19.6
LY = PQ - MQ = 6.4
NY = LY - MQ = 2.8
XY^2 = NX^2 + NY^2 = 392
