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# A triangle has 3 vertices (1,2), (3,-4), and (-2,3). Find

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A triangle has 3 vertices (1,2), (3,-4), and (-2,3). Find [#permalink]  18 Jun 2006, 11:07
A triangle has 3 vertices (1,2), (3,-4), and (-2,3). Find the area of the triangle.

I found a solution using the determinant route from another site but I could not follow it at all. Please help shed light on this puzzler.

-Tru
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[#permalink]  18 Jun 2006, 12:04
Plot it on an XY plane, calculate each side (X2-X1), (X3-X2) etc.

Once you have 3 sides (a,b,c), then the area is calculated using

A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2

Often, these questions can be far simpler too. When you plot it on the XY plane, see if you can get the height and the base (from any orientation), if you have that - just do

(1/2)*base*height

for the answer.

The key here is to do a quick plot of the triangle on an XY plane.
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[#permalink]  18 Jun 2006, 23:18
In such a case, the answers might help. Aren't there any choices? What is the source of the question?

The semi-perimeter s = 4 + 1.5sqrt(10)
The area is sqrt (6.5*13.75) using the formula sqrt[s(s-a)(s-b)(s-c)]

I am not sure if such complex calculations would be asked in the GMAT. Plotting the triangle helps to an extent. However with the given coordinates altitudes and bases are difficult to detemine.
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[#permalink]  18 Jun 2006, 23:41
I plotted the triangle. Can solve by finding area of various triangles and rectangles.

Area come to 4 sq units
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[#permalink]  19 Jun 2006, 08:36
OA is 8 square units.

I remember doing a question from the GMAT Prep test involving vertices, which flustered me. So I went looking for help. I found the example given in this thread on a math forum while looking for that help.

I found out how to solve the problem using 3x3 determinants but it is to confusing and hard to share at this moment. Perhaps I can show the technique using illustrator when I have time later this evening.

Thanks for the input thus far.

-Tru
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[#permalink]  19 Jun 2006, 09:26
Yes, the answer is 8 and I got it by using a really simple technique:
what I did was just drawing a triangle and then inscribing it inside a rectangle with the vertices (-2;3)(3;3)(3;-4)(-2;-4). Now, it's pretty simple to find the area of the rectangle with sides 5 and 7, it's 35. Then I just substructed from this the 'extra' bits and pieces (three right triangles and a rectangle): 35-2-6-1.5-17.5=8
Huh?
[#permalink] 19 Jun 2006, 09:26
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# A triangle has 3 vertices (1,2), (3,-4), and (-2,3). Find

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