Join diagonal AC of quadrilateral ABCD.
Here, the sides of triangle ABC are a = 3 cm, b = 4 cn and c = 5 cm.
So, the semi-perimeter of triangle S = a+b+c/2 = 3 + 4 + 5/2 = 12/2 = 6 cm
Therefore, using Heron’s formula area of triangle = √s(s -a)(s -b)(s -c)
= √6(6 -3)(6 -4)(6 -5) = √6(3)(2)(1) = √36 = 6 cm²
And the sides of triangle ACD are a’ = 4 cm, b’ = 5 cn and c’ = 5 cm.
So, the semi-perimeter of triangle S’ = a’ + b’ + c’/2 = 4 + 5 + 5/2 = 14/2 = 7 cm
Therefore, using Heron’s formula area of triangle = √s'(s’ – a’)(s’ -b’)(s’ -c’)
= √7(7 – 4)(7 – 5)(7 – 5)
= √7(3)(2)(2)
= 2√21 = 9.2 cm²(approx.)
Total area of quadrilateral = Area of triangle ABC + Area of triangle ACD
⇒ Total area of quadrilateral ABCD = 6 + 9.2 = 15.2 cm²
Hence, the area of quadrilateral ABCD is 15.2 cm².