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Find the area of this parallelogram. D(-3,4) c(3,4) a(-6,-4) b(0,-4)

Find The Area Of This Parallelogram D34 C34 A64 B04 class=

Sagot :

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.

Learn more about area at https://brainly.com/question/25965491

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