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Question

Draw a quadrilateral in the Cartesian plane, whose vertices are (4,5),(0,7),(5,5) and (4,2). Also, find its area.

Solution
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A(ABCD)=A(ABC)+A(ACD)

(4,5)(x1,y1),(0,7)(x2,y2),(5,5)(x3,y3)

Now, area of ABC

=12|x1(y2y3)+x2(y3y2)+x3(y1y2)|

=12|4(7+5)+0(55)+5(57)|

=12|4(12)+5(2)|

=12|4810|

=12|58|

=12×58

=29 sq.units

Area of ACD

=12|4(5+2)+5(25)+(4)(5+5)|

=12|4(3)+5(7)4(10)|

=12|123540|

=12|63|

=632 sq.units

So, area (ABCD)=(29+632)

=58+632

=1212 sq.units

404357_418290_ans_f269642c4dd942149ee94f8b3285e9f1.png

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