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.