Question

find the area of quadrilateral abcd whose vertices are A(1,2),B(6,2),C(5,3),D(3,4)

Answer 1

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Answer 2

$A(1,2), B(6,2), C(5,3) \:and \: D(3,4) \:are \\vertices \: of \: a \: Quadrilateral \: ABCD$

$Join \: A \: to \: C$

$i ) Let \: A(1,2) = (x_{1},y_{1}) ,\\ B(6,2) = (x_{2},y_{2}) \: and \: C(5,3) = (x_{3},y_{3})$

$Area \:of \: triangle \:ABC \\= \frac{1}{2}|x_{1}(y_{2} - y_{1}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) |$

$= \frac{1}{2}|1(2-3) + 6(3-2)+5(2-2)|\\= \frac{1}{2}|-1+6+0| \\= \frac{1}{2}|5| \\= \frac{5}{2} \: ---(1)$

$ii ) Let \: A(1,2) = (x_{1},y_{1}) ,\\ D(3,4) = (x_{2},y_{2}) \: and \: C(5,3) = (x_{3},y_{3})$

$Area \:of \: \triangle \:ADC \\= \frac{1}{2}|x_{1}(y_{2} - y_{1}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) |$

$= \frac{1}{2}|1(4-3) + 3(3-2)+5(2-4)|\\= \frac{1}{2}|1+3-10| \\= \frac{1}{2}|-6| \\= \frac{6}{2} \: ---(2)$

Therefore.,

$\red{ Area \:of \:the \: Quadrilateral } \\\green { =area(\triangle {ABC}) + area(\triangle {ADC})}\\= \frac{5}{2} + \frac{6}{2} \\= \frac{ 5 + 6}{2} \\= \frac{11}{2} \\= 5.5 \:cm^{2}$

$\red{Area \: of \: the \: ABCD \: Quadrilateral }\green {= \frac{11}{2} \:cm^{2} }$

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