Consider △ BCD
Using the Pythagoras theorem
We can write it as
DB2 + BC2 = DC2
By substituting the values
DB2 + 82 = 172
By subtraction
DB2 = 172 – 82
DB2 = 289 – 64
By subtraction
DB2 = 225
By taking the square root
DB = √ 225
So we get
DB = 15cm
We can find
Area of △ BCD = ½ × b × h
By substituting the values
Area of △ BCD = ½ × 8 × 15
On further calculation
Area of △ BCD = 60 cm2
Consider △ BAD
Using the Pythagoras theorem
We can write it as
DA2 + AB2 = DB2
By substituting the values
AB2 + 92 = 152
By subtraction
AB2 = 152 – 92
AB2 = 225 – 81
By subtraction
AB2 = 144
By taking the square root
AB = √ 144
So we get
AB = 12cm
We can find
Area of △ DAB = ½ × b × h
By substituting the values
Area of △ DAB = ½ × 9 × 12
On further calculation
Area of △ DAB = 54 cm2
So we get
Area of quadrilateral ABCD = area of △ DAB + area of △ BCD
By substituting the values
Area of quadrilateral ABCD = 54 + 60
By addition
Area of quadrilateral ABCD = 114 cm2
Therefore, the area of quadrilateral ABCD is 114 cm2.