Find the area of a quadrilateral ABCD in which AB = 3 cm , BC = 4 cm , CD = 6 cm , AD = 5 cm and diagonal AC = 5 cm . note - answer is 18 proof it you will be marked brainleast.
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2S=5+4+3 ⇒ S=6
Area of ΔABC
=
s(s−a)(s−b)(s−c)
=
15(15−9)(15−8)(15−13)
=
6(1)(2)(3)
=
36
=6
⇒For area of ΔADC,
⇒2S=5+4+5
⇒ S=7cm
⇒Area of ΔADC
=
7(7−5)(7−5)(7−4)
=
7×2×2×3
=2
21
=9.16m
2
⇒Area of quad ABCD= Area of (△ABC+△ADC)
=6+9.16
=15.16 cm
2
≈15.2 cm
2
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