Since the quadrilateral is cyclic,
#color(red)(/_BCD) = 180^circ-/_DAB = 80^circ#
Using the cosine law for the triangle #Delta ABD# we get
#BD^2 = AB^2+AD^2-2 times AB times AD cos /_DAB#
# qquad = (7.2^2+6.1^2-2times 7.2times 6.1 times cos 100^circ)" cm"^2#
#qquad = 104.303" cm"^2#
Thus, #color(blue)(BD= 10.2 " cm")#
Applying the sine law to #Delta ABD# we further get
#{sin/_ABD}/{AD} = {sin/_BDA }/{AB} = {sin/_DAB}/BD qquad implies#
#{sin/_ABD}/{6.1" cm"} = {sin/_BDA }/{7.2 " cm"} = {sin100^circ}/{10.2" cm"} = 0.0964" cm"^-1#
Thus
#sin/_ABD = 0.5882 impliescolor(blue)( /_ABD = 36^circ)#
and
#color(blue)(/_BDA)= 180^circ-100^circ - 36^circ = color(blue)(44^circ)#
Applying the sine law again, but this time to #Delta BCD#, we get
#{sin/_DBC}/{CD} = {sin/_CDB }/{BC} = {sin/_BCD}/{BD} qquad implies#
#sin/_CDB = {BC}/{BD}sin/_BCD = (8.2" cm")/(10.2" cm")sin 80^circ = 0.7907#
#implies color(blue)(/_CDB )= sin^-1(0.7907) = color(blue)(52.5^circ)#
and hence
#color(blue)(/_DBC) = 180^circ-80^circ-52.5^circ = color(blue)(47.5^circ)#
So,
#color(red)(/_ADC) = /_BDA + /_CDB = 44^circ+52.5^circ = color(red)(96.5^circ)#
Now, we also have
# color(blue)(CD) = {sin /_DBC)/(sin/_BCD) BD = {sin 47.5^circ}/{sin80^circ} times 10.2" cm" = color(blue)(7.64" cm")#
So, the perimeter is
#AB+BC+CD+DA = 7.2" cm" + 8.2 " cm"+7.6" cm"+6.1" cm" = color(red)(29.1" cm")#
The area of the triangle #Delta ABD# is
#1/2 times AB times AD times sin/_DAB = 1/2times 7.2times6.1times sin100^circ" cm"^2 = color(blue)(21.63" cm"^2)#
while that of #Delta BCD# is
#1/2 times BC times CD times sin/_BCD = 1/2times 8.2times7.6times sin80^circ" cm"^2 = color(blue)(30.69" cm"^2)#
So, the area of the cyclic quadrilateral is #(21.63+30.69)" cm"^2 = color(red)(52.3" cm"^2)#
