Question 274213: I don't know how to do this.
Directions: The vertices of an irregular figure are given. Find the area of the figure.
T(-4,-2), U(-2,2), V(3,4), W(3,-2)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The vertices of an irregular figure are given. Find the area of the figure.
T(-4,-2), U(-2,2), V(3,4), W(3,-2)
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There's more than one way.
The most obvious is to make it 2 triangles, solve for the lengths and find the areas.
A better way is:
....T. .U. .V. .W ..T
x -4 -2 +3 +3 -4
y -2 +2 +4 -2 -2
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Multiply the elements with a slope of -1, ie, Tx*Uy, Ux*Vy, etc and add them
Multiply the elements with a slope of +1, ie, Ty*Ux, Uy*Vx, etc and add them
The absolute value of the difference is 2 x the area, divide by 2.
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(-4*2) + (-2*4) + (3*-2) + (3*-2) = -8 - 8 - 6 - 6 = -28
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((-2*-2) + (2*3) + (4*3) + (-2*-4) = 4 + 6 + 12 + 8 = 30
Difference = 58
Area = 29 sq units
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This method is good for any number of points. Note that the 1st and last points are the same.
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