So ∠ ABC+∠ ADC = 180, connecting communication, by cosine theorem:
AC2=42+62-2×4×6×cos∠ABC
=42+22-2×2×4cos∠ADC,
So cos∠ABC= 12, ∫∠ABC ∈( 0, π),
Therefore ∠ ABC = 60.
S quadrilateral ABCD =12× 4× 6× sin60+12× 2× 4× sin120.
=83 (ten thousand square meters).
In △ABC, through cosine theorem:
AC2=AB2+BC2-2AB? BC? cos∠ABC
= 16+36-2×4×6× 12.
AC=27。
According to sine theorem asinA=bsinB=2R,
∴2R=ACsin∠ABC=2732=42 13,
∴R=22 13 (ten thousand meters).
(2)∫S quadrilateral APCD=S△ADC+S△APC,
S△ADC= 12AD? CD? sin 120 =23,
Let AP=x and CP = y.
So S△APC= 12xy? sin60 =34xy。
And cosine theorem ac2 = x2+y2-2xycos 60.
=x2+y2-xy=28。
∴x2+y2-xy≥2xy-xy=xy.
∴xy≤28, equals sign if and only if x = y.
∴S quadrilateral apcd = 23+34xy ≤ 23+34x28 = 93,
The largest area is 930,000 square meters.