Find the area of a quadrilateral ABCD in which AB=3 cm, BC=4 cm, CD=4 cm, DA=5 cm and AC=5 cm

2S=5+4+3⇒S=6
Area of ΔABC=√s(s−a)(s−b)(s−c)=√6(6−5)(6−4)(6−3)=√6(1)(2)(3)=√36=6
⇒For area of ΔADC,⇒2S=5+4+5⇒S=7cm
⇒Area of ΔADC=√7(7−5)(7−4)(7−5)=√7×2×3×2=2√21=9.16m2
⇒Area of quadABCD=Area of(ΔABC+ΔADC)=6+9.16=15.16 cm2≈15.2 cm2