find the area of the quadrilateral abcd whose vertices are a(3 -1) b(9 -5) c(14 0) and d(9 19)
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Join A and CNow, Area of quadrilateral ABCD = Area of\xa0{tex} \\Delta{/tex}ABC + Area of\xa0{tex}\\Delta{/tex}ACDArea of\xa0{tex} \\Delta A B C = \\frac { 1 } { 2 } | 3 ( – 5 – 0 ) + 9 ( 0 + 1 ) + 14 ( – 1 + 5 ) |{/tex}{tex}= \\frac { 1 } { 2 } | – 15 + 9 + 56 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 50{/tex}= 25 sq. unitsArea of\xa0{tex}\\Delta ACD = \\frac{1}{2}|3(0 – 19) + 14(19 + 1) + 9( – 1 – 0)|{/tex}{tex}= \\frac { 1 } { 2 } | – 57 + 280 – 9 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 214{/tex}= 107 sq. unitsArea of quad. ABCD = Area of\xa0{tex}\\Delta{/tex}ABC + Area of\xa0{tex}\\Delta{/tex}ACD= (25 + 107) sq. units=132 sq. units