find the area of a quadrilateral whose vertices are (3,5),(6,6),(4,3)and (1,2)
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Step-by-step explanation:
Area of a quadrilateralABCD=Area of △ABD+Area of △BCD
Area of a triangle=
2
1
∣
∣
∣
∣
∣
∣
∣
∣
x
1
x
2
x
3
y
1
y
2
y
3
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
Area of a △ABD=
2
1
∣
∣
∣
∣
∣
∣
∣
∣
−3
−5
1
4
−6
2
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=
2
1
[−3(−6−2)−4(−5−1)+1(−10+6)]
=
2
1
[24+24−4]=
2
44
=22
Area of a △BCD=
2
1
∣
∣
∣
∣
∣
∣
∣
∣
−5
4
1
−6
−1
2
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=
2
1
[−5(−1−2)+6(4−1)+1(8+1)]
=
2
1
[15+18+9]=
2
42
=21
∴ Area of a quadrilateralABCD=Area of △ABD+Area of △BCD=22+21=43sq.units.
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