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sri340
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#1
mathcounts
triangle verticis problem

Triangle ABC has its vertices at A(0,4), B(0,9) and C(6,9). What is the area of the triangle, in square units?

PostPosted: Fri Nov 06, 2009 3:11 am  Back to top 
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gauss1181
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#2
If you draw out the triangle on the x-y coordinate plane, you'll see it has a base of 5 and a height of 9. So the area is just \frac{1}{2}(5)(9)=\boxed{22.5} square units.
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PostPosted: Fri Nov 06, 2009 7:24 am  Back to top 
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monkeygirl13
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#3
Here is a much faster way to do the problem.
Solution

The difference between 0,9 and 0,4 is 5.
The difference between 6,9 and 0,9 is 6.
To get the area of this triangle you solve the expression
\frac{1}{2}(5)(6)= 15
Note: I am a very unsmarticle person pm if I made a mistake.

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PostPosted: Fri Nov 06, 2009 3:28 pm  Back to top 
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randomguy64
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#4
I agree with monkeygirl.

PostPosted: Fri Nov 06, 2009 5:05 pm  Back to top 
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Stank
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#5
Monkeygirl - Your solution works just fine; however, it's important to graph it (or at least look at the slopes of the lines). For all you know, they could have made it some irregular triangle and you would have needed to use Herron's Formula, Shoelace Formula, etc.
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PostPosted: Fri Nov 06, 2009 7:00 pm  Back to top 
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randomguy64
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#6
You realize that (0, 9) and (0, 4) share the same x value, and (6, 9) and (0. 9) share the same y-value, making the triangle right and half a rectangle that is "parallel" to the axes...

PostPosted: Fri Nov 06, 2009 7:44 pm  Back to top 
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mathkid95
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#7
Well, You could plot the points out and use the distance formula for the base and height and multiply that by \frac {1}{2}

So, it is \sqrt {(x_2 - x_1)^{2} + (y_1 - y_ 2)^{2}}

Plug in what you know and you get the distances between the base: BC and the height: AB to get

\frac {1}{2}(5)(6) = 15

The area is 15 square units
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PostPosted: Sat Nov 07, 2009 8:19 am  Back to top 
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