Question

Find the area of a polygon with the given vertices? A(1, 4), B(-2, -2) C(-7, -2), D(-4, 4) Please show work.

Answer 1

`S=30`

Explanation:

Consider that the polygon ABCD is composed of the triangle ABC and ACD.

To find the area of a triangle whose vertices coordinates are given we can use the Cramer's Rule, described in:
Finding the area of a triangle using the determinant of a matrix

Evaluating the determinant of the Cramer's Rule we get:
`S_(triangle) =(1/2)|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|`
`S_(triangle)=(1/2)|x_1*(y_2-y_3)+x_2*(y_3-y_1)+x_3*(y_1-y_2)|`

For `triangle`ABC
A(1, 4)
B(-2, -2)
C(-7, -2)
`S_(triangleABC)=(1/2)|1*(-2+2)+(-2)(-2-4)+(-7)(4+2)|`
`S_(triangleABC)=(1/2)|0+12-42|=(1/2)*30=15`

For `triangle`ACD
A(1, 4)
C(-7, -2)
D(-4, 4)
`S_(triangleACD)=(1/2)|1*(-2-4)+(-7)(4-4)+(-4)(4+2)|`
`S_(triangleACD)=(1/2)|-6+0-24|=(1/2)*30=15`

`S_(ABCD) = S_(triangleABC)+S_(triangleACD)=15+15=30`

Answer 2

S=30

Explanation:

Repeating the points
A(1,4)
B(-2,-2)
C(-7,-2)
D(-4,4)

If we plot those points we'll see that A and D are in the same line (`y=4`) parallel to the x-axis and that B and C also are in the same line (`y=-2`) also parallel to the x-axis.

Beyond that, since A and D are in the same line and also B and C are in the same line
`DA=|x_A-x_D|=|1+4|=5`
`BC=|x_B-x_B|=|-7+2|=5`
=> `DA=BC`

Two segments of line of the same size in lines parallel to each other, yet the segments are not aligned: it means that the polygon is a parallelogram, whose equation of area is `base*height`.

The separation or distance between the two lines (`y=4` and `y=-2`) give us the height. The separation is `4-(-2)=6` linear units.

So the area of the polygon ABCD, a parallelogram, is
`S_(ABCD)=base*height=5*6=30`