Now, apply Pythagorean theorem in ΔABC,

AC² = AB² + BC²

⇒ 5² = 3²+4²

⇒ 25 = 25

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (1/2 × 3 × 4) = 6 cm²

The semi perimeter of ΔACD (s) = (perimeter/2)

= (5+5+4)/2 cm = 14/2 cm

= 7 m

Now, using Heron’s formula,

Area of ΔABD

= \(\sqrt{s(s - a)(s - b)(s - c)}\)

= \(\sqrt{7(7 - 5)(7 - 5)(7 - 4)}\)cm²

= \(\sqrt{7 \times 2 \times 2 \times 3}\)cm²

= 2√21 cm² 

= 9.17 cm² (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD

= 6 cm² + 9.17 cm² 

= 15.17 cm²