So angle A+ angle C= 180 degrees,
So COSA =-COSC,
According to cosine theorem, BD 2 = AB 2+DA 2-2 xadxdaxosa.
bd^2=bc^2+cd^2-2xbcxcdxcos,
So1+4-4 COSA = 4+9-12 COSC.
5+4 Xhosa = 13- 12 Xhosa
16cosC=8
cosC= 1/2
Angle C=60 degrees.
So BD 2 = 4+9- 12x (1/2).
=7
BD= root number 7.
(2) Because the angle C=60 degrees,
So the angle A= 120 degrees,
So the area of the triangle ABD = (1/2) xabxasina.
=( 1/2)x 1x2x (root number 3)/2
= (root number 3)/2,
Area of triangle BCD =( 1/2)xBCxCDxsinC
=( 1/2)x2x3x (root number 3)/2
=(3 radicals 3)/2,
So the area of quadrilateral ABCD = the area of triangle ABD+the area of triangle BCD.
The root number 3 of =2.
According to sine theorem:
Circumferential circle diameter =BD/sinC
= (root number 7)/[ (root number 3)/2]
=(2 radicals 2 1)/3.