Find the area of a parallelogram PGRM with vertices at (0,0) (6,0) (2,4) and (8,4)

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Respuesta :

frika frika · Answer 1 · 2019-06-11T10:38:15+02:00

Answer:

[tex]A=24\ un^2.[/tex]

Step-by-step explanation:

Plot points A(0,0), B(6,0), C(2,4) and D(8,4) on the coordinate plane (see attached diagram). The segment CE is the height of the parallelogram ABDC.

The area of the parallelogram is

[tex]A=\text{Base}\cdot \text{Height}[/tex]

Base= AB

Height =CE

So,

[tex]AB=\sqrt{(6-0)^2+(0-0)^2}=\sqrt{36+0}=\sqrt{36}=6\\ \\CE=\sqrt{(2-2)^2+(4-0)^2}=\sqrt{0+16}=\sqrt{16}=4[/tex]

Hence, the area of the parallelogram is

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DeanR DeanR · Answer 2 · 2019-06-11T10:40:36+02:00

I'll tell you how to do it for any polygon in the cartesian plane with the vertices listed in order.

First we have to list the vertices in order so each pair is a side:

(0,0) (6,0) (8,4) (2,4)

Now for each side (a,b)(c,d) we calculate the cross product ad-bc

(0,0)(6,0) 0(0)-0(6)=0

(6,0)(8,4) 6(4)-0(8)=24

(8,4)(2,4) 8(4)-4(2) = 24

(2,4)(0,0) 2(0)-4(0)=0

We add up the cross products, and take half the absolute value of the sum for the area:

Area = (1/2) | 0 + 24 + 24 + 0 | = 24

Answer: 24