Answer :

For triangle ABC

Area of triangle

= 1/2(x_{1}(y_{2}−y_{3}) + x_{2}(y_{3}−y_{1}) + x_{3}(y_{1}−y_{2}))

= 1/2(3(–5–0) + 9(0 + 1) + 14(–1 + 5))

= 1/2(–15 + 9 + 56)

= 1/2(50)

= 25

For triangle ACD

Area of triangle

= 1/2(x_{1}(y_{2}−y_{3}) + x_{2}(y_{3}−y_{1}) + x_{3}(y_{1}−y_{2}))

= 1/2(3(0–19) + 14(19 + 1) + 9(–1–0))

= 1/2(–57 + 280–9)

= 1/2(214)

= 107

Area of ABCD = Area of ABC + Area of ACD

= 25 + 107

= 132 sq units

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