SOLUTION: What are the areas of a rectangle whose vertices are (-2,5), (8,5), (8-2) and (-2,-2)?

Question 877238: What are the areas of a rectangle whose vertices are (-2,5), (8,5), (8-2) and (-2,-2)?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Do it like this one:
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What is the area of a polygon with vertices (-2,2), (3,2), (7,-5), and (-2,-5)?
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Add labels
A(-2,2), B(3,2), C(7,-5), and D(-2,-5)
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List them in a 2 by -- matrix & repeat the 1st point
A...B...C...D...A
-2..3...7..-2..-2
2...2..-5...-5..2
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Add the diagonal products starting at the upper left
-2*2 + 3*-5 + 7*-5 + -2*2 = -4 - 15 - 35 - 4 = -58
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Add the diagonal products starting at the lower left
2*3 + 2*7 + -5*-2 + -5*-2 = 6 + 14 + 10 + 10 = 40
Get the absolute value of the difference = 98
The area is 1/2 that = 49 sq units
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Works for any # of points, but they must be in order on the perimeter.
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