A park in the shape of a quadrilateral ABCD,has angleC=90degree,AB=9m,BC=12cm,CD=5m and AD=8m.How much area do

A park in the shape of a quadrilateral ABCD,has angleC=90degree,AB=9m,BC=12cm,CD=5m and AD=8m.How much area does it occupy?

Join AC, then
AC=sqrt(8^2+5^2)=>
AC=sqrt(89) =>
AC=9.434 m approximately.
Let s=(9+12+9.434)/2=15.217 m
The area=sqrt[s(s-9)(s-12)(s-9.434)]
area=sqrt[15.217(6.217)(3.217)(5.783)]…
area=41.952 sq.m
approximately.

as per the data only angle C is 90 deg
Join BD
area triangle BCD= (1/2)*12*5=30 sq m
for area triangle ABD
BD=13 as it is a hypotenuse
area=(1/)*sqrt{ P*(P-2a)*(P-2b)*(P-2d) }
P=perimeter=30
a=13,b=8,d=9
we get area=(1/4)*sqrt(30*14*12*4)
=35.50 sq m
total area=BCD+ADB
=65.50 sq m

I guess you mean BC = 12 m. Then it's 10√65 m²