SOLUTION: Determine the area of the right triangle with vertices R(4,4), S(-2,-2), and T(10,-2).

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Question 1159866: Determine the area of the right triangle with vertices R(4,4), S(-2,-2), and T(10,-2).
Found 2 solutions by greenestamps, Alan3354:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The y-coordinates of S and T are the same, so side ST is horizontal, with length 10-(-2) = 12.

So we can use ST as the base; the height of the triangle is then just the distance from the line y=-2 to the point (4,4), which is 4-(-2) = 4+2=6.

Then the area is one-half base times height: %281%2F2%29%2812%29%286%29+=+36

Alternatively (but I think it is more work), you could use the fact that it is a right triangle, with legs RS and RT, so the area is one-half the product of the lengths of those legs.

The length of each leg is 6sqrt%282%29; then the area is %281%2F2%29%286sqrt%282%29%29%286sqrt%282%29%29+=+%281%2F2%29%2872%29+=+36

But that, like I said, was more work....

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the area of the right triangle with vertices R(4,4), S(-2,-2), and T(10,-2).
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Another method:
   R   S   T   R
x  4  -2  10   4
y  4  -2  -2   4

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Add the diagonal products starting at the upper left:
--> 4*(-2) + -2*-2 + 10*4 = -8 + 4 + 40 = 36
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Add the diagonal products starting at the lower left:
--> 4*-2 + -2*10 + -2*4 = -8 -20 - 8 = -36
The difference is 72.
The area is 1/2 that, = 36 sq units.
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Might be the "hard way" for a triangle, but it works for any plane figure, with any # of sides.
The points must be entered in order around the figure is the only restriction.