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| 1. |
Area of a quadrilateral ABCD whose vertices are A(3,-1),B(9,-5),C(14,0),D(9,19) |
| Answer» 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 | |