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
Open in App
Solution
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=6cm2⇒For area of △ADC,⇒2s=5+4+5⇒S=7cm⇒Areaof△ADC=√s(s−a)(s−b)(s−c)=√7(7−5)(7−5)(7−4)=√7×2×2×3√221=9.16cm2⇒Area of quad ABCD=Area of(△ABC+△ADC)
= 6 + 9.16 =15.16cm2≈15.2cm2