Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B (2 + 
, 5) and C(2, 6).
It is given that A(2, 4), B(2 + 
, 5) and C(2, 6) are the vertices of the parallelogram ABCD.
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area of □ABCD = 2 × Area of ∆ABC
Area of ∆ABC = 
|2(5 - 6)+ (2 + 
)(6 - 4)+2(4 - 5)|
= 
|-2 + 4 + 
- 2|
= 
× 
= 
sq. units
∴ Area of □ABCD = 2 × 
= 
sq. units
Hence, the area of given parallelogram is 
sq. units
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