Answer:
[tex]222cm^{2}[/tex]
Step-by-step explanation:
First we will find the length of BD.
By Pythagoras' Theorem,
[tex]c^{2} =a^{2} +b^{2} \\BD^{2} = AD^{2} +AB^{2} \\BD^{2} =12^{2} +16^{2} \\BD^{2} = 144+256\\BD^{2} =400\\BD = \sqrt{400} \\=20cm[/tex]
Now we will find Angle ADB.
[tex]\frac{AB}{sin(Angle ADB)} =\frac{BD}{Sin(AngleBAD)} \\\frac{16}{sin(AngleADB)} =\frac{20}{sin(90)} \\20sin(AngleADB)=16sin(90)\\20sin(AngleADB)=16\\sin(AngleADB)=\frac{16}{20} \\sin(AngleADB)=0.8\\AngleADB=sin^-1(0.8)\\=53.130deg\\[/tex]
Now we will find Area of triangle BCD
[tex]Angle BDC = 90 - Angle ADB = 90 - 53.130 = 36.87deg\\\\\\Area of Triangle BCD \\= \frac{1}{2} (BD)(CD)sin(AngleBDC)\\=\frac{1}{2} (20)(21)sin(36.87)\\= 126cm^{2}[/tex]
Then we will find the area of triangle BAD
[tex]=\frac{1}{2} (16)(12)\\= 96cm^{2} \\[/tex]
Area of quadrilateral = Area of Triangle BCD + Area of Triangle BAD
= 126+96
= [tex]222cm^{2}[/tex]
