Find the area of quadrilateral ABCD …
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Sia ? 6 years, 6 months ago
Area of quadrilateral ABCD = ar {tex}\triangle{/tex}ABD + ar {tex}\triangle{/tex}BCD
ar {tex}\triangle{/tex}ABD = {tex}\frac 12{/tex}[ 1(- 3 - 21) + 7(21 - 1) + 7(1 + 3)]
= {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} = 72 sq. units
ar {tex}\triangle{/tex}BCD
= {tex}\frac { 1 } { 2 }{/tex}[7(2 - 21) + 12(21 + 3) + 7(-3 - 2)]
= {tex}\frac { 1 } { 2 } [ 7 \times - 19 + 12 \times 24 + 7 \times - 5 ]{/tex}
= {tex}\frac12{/tex}[-133 + 288 - 35]
= {tex}\frac 12{/tex}[288 - 168]
{tex}= \frac { 1 } { 2 } \times 120{/tex}
= 60 sq. units
Hence, Area Quadrilateral ABCD = 72 + 60 = 132 sq units
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