Question 512940: Find the area of the triangle whose vertices have coordinates (-8,3), (-8,-4), and (6,3).
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the area of the triangle whose vertices have coordinates (-8,3), (-8,-4), and (6,3).
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Label A(-8,3), B(-8,-4), and C(6,3).
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A & B have the same x value, call AB the base, length = 7
A & C have the same y value, so it's a right triangle, height = 14
Area = b*h/2 = 49 sq units.
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A general method:
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A.. B.. C.. A
-8 -8 +6 -8
+3 -4 +3 +3
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Add the diagonal products starting upper left
-8*-4 + -8*3 + 6*3 = 32 - 24 + 18 = 26
Add the diagonal products starting lower left
3*-8 + -4*6 + 3*-8 = -24 - 24 - 24 = -72
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The difference is 98.
The area is 1/2 that, = 49 sq units.
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Same answer, but the 2nd method works for ANY polygon, ANY # of sides.
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