Find the area of the parallelogram ABCD three of its vertices are (1,2,3), (2,3,1) and (3,1,2)
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Step-by-step explanation:
We know that the diagonal of a parallelogram divides the parallelogram into 2 equal triangles.
So, Area of parallelogram,
ABCD=2× area of ΔABC ......... (1)
We know that,
Area of ΔABC with vertices (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
) is
A=
2
1
×[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
So, Area of ΔABC with vertices [A(2,4),B(2+
3
,5),C(2,6)] is
A=
2
1
×[2(5−6)+(2+
3
)(6−4)+2(4−5)]
A=
2
1
×[−2+2(2+
3
)−2]
A=
2
1
×[−2+4+2
3
−2]
A=
3
Put this value in equation (1), we get,
Area of parallelogram,
ABCD=2× area of ΔABC
ABCD=2
3
sq units
Hence, area of the parallelogram is 2
3
sq units.
December 27, 2019
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