Given:

AC = 15 cm

AB = 17 cm

AD = 9 cm

CD = 12 cm

In ∆ACB (right-angled),

Base² + Perpendicular² = Hypotenuse²

⇒ BC² + AC² = AB²

⇒ BC² = AB² - AC²

⇒ BC² = 17² - 15²

⇒ BC² = 289 - 225

⇒ BC² = 64

⇒ BC= 8 cm

Area of ∆ACB = 1/2 × BC × AC

= 1/2 × 8 cm × 15 cm

= 60 cm²

In ∆ADC,

Area of ∆ADC = 1/2 × AD × CD

= 1/2 × 9 cm × 12 cm

= 54 cm²

Now,

Area of quadrilateral ABCD = Area of ∆ACB + Area of ∆ADC

= 60 cm² + 54 cm²

= 114 cm²

And,

Perimeter of quadrilateral ABCD = AB + BC + CD + DA

= 17 cm + 8 cm + 12 cm + 9 cm

= 46 cm