Question 615209: What is the area of a square with the vertices (3,3),(6,6),(9,3)and (6,0)
A 3 square root of 2 units
B 12 square root of 2 units^2
C 18 units^2
D 36 units^2
this problem has been giving me the biggest headache if you would be so kind to send me the steps of how you solved it i would be very grateful
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! What is the area of a square with the vertices (3,3),(6,6),(9,3)and (6,0)
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Call the vertices A, B, C & D
Arrange them:
A B C D A
3 6 9 6 3
3 6 3 0 3
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Add the diagonal products starting with the upper left
= 3*6 + 6*3 + 9*0 + 6*3 = 54
Add the diagonal products starting with the lower left
3*6 + 6*9 + 3*6 + 0*3 = 90
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The area = 1/2 the absolute value of the difference
= (1/2)*36
Area = 18
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This works for all polygons, any # of sides.
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