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| 1. |
Find the area of quadrilateral ABCD whose vertices are A (1,1)B (7,-3)C(12,2)and D(7,21). |
| Answer» Area of quadrilateral ABCD = ar {tex}\\triangle{/tex}ABD + ar {tex}\\triangle{/tex}BCDar {tex}\\triangle{/tex}ABD =\xa0{tex}\\frac 12{/tex}[ 1(- 3 - 21) + 7(21 - 1) + 7(1 + 3)]=\xa0{tex} \\frac { 1 } { 2 } [ - 24 + 7 \\times 20 + 7 \\times 4 ]{/tex}{tex}= \\frac { 1 } { 2 } {/tex}[-24 + 140 + 28]{tex}= \\frac { 1 } { 2 } \\times 144{/tex}\xa0= 72 sq. unitsar {tex}\\triangle{/tex}BCD=\xa0{tex}\\frac { 1 } { 2 }{/tex}[7(2 - 21) + 12(21 + 3) + 7(-3 - 2)]=\xa0{tex}\\frac { 1 } { 2 } [ 7 \\times - 19 + 12 \\times 24 + 7 \\times - 5 ]{/tex}=\xa0{tex}\\frac12{/tex}[-133 + 288 - 35]=\xa0{tex}\\frac 12{/tex}[288 - 168]{tex}= \\frac { 1 } { 2 } \\times 120{/tex}= 60 sq. unitsHence, Area Quadrilateral ABCD = 72 + 60 = 132 sq units | |