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.

2 Answers
Jan 1, 2016

Answer:

#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#

Jan 2, 2016

Answer:

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#