From cosine theorem: △ABC, | AC | 2 = 4 2+6 2-2 * 4 * 6 * COSB.
△ In △ACD，| AC | 2 = 4 ^ 2+2 ^ 2-2 * 4 * 2 *( 180-b)。
Combining the above two formulas, we can get B = 60, so D = 120, | AC | = 2 √ 7.
△ ABC = | AB | *| BC | SINB/2 = (4 * 6 * SIN60)/2 = 6 √ 3 area.
Similarly, the area of delta △ACD =2√3.
Then the area of the quadrilateral ABCD =S△ABC+S△ACD=8√3.
2) In △ABC, the sine theorem is |AC|/sinB=2R, so the radius of the circumscribed circle of the quadrilateral is R=2√2 1/3.
3) Connect AC, the area of quadrilateral APCD =S△APC+S△ACD, and the area of △ACD is a fixed value of 2√3.
In order to maximize the area of △APC and AC to a constant value, it is necessary to maximize the height, that is, the distance from point P to AC is maximized.
At this time, point P is the midpoint of the arc ABC, and PA=PC. Since ∠ P = 60, δ△APC is positive δ.
The area is 7√3. At this time, the area of quadrangle APCD is the largest, which is 9√3.