If the vertices of parallelogram ABCD is A(-6,-4),B(0,-4),C(3,4),D(-3,4) ,then the area of parallelogram ABCD will be 48 square units.
Given that the vertices of parallelogram ABCD is A(-6,-4),B(0,-4)
,C(3,4),D(-3,4).
We are required to find the area of parallelogram ABCD.
Let the point of intersection of height of parallelogram from point A is point E.
When we see the graph on which the parallelogram then we will be able to know that point E has coordinates of (-3,-4).
Area of parallelogram=Base*Height
=AB*DE
DE=[tex]\sqrt{(-4-4)^{2} +(-3+3)^{2} }[/tex]
=[tex]\sqrt{64}[/tex]
=8 units
AB=[tex]\sqrt{(-4+4)^{2} +(0+6)^{2} }[/tex]
=[tex]\sqrt{36}[/tex]
=6 units
Area=6*8
=48 square units.
Hence if the vertices of parallelogram ABCD is A(-6,-4),B(0,-4),C(3,4),D(-3,4) ,then the area of parallelogram ABCD will be 48 square units.
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