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In the rectangular coordinate system, the vertices of a triangle have

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In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 01 Mar 2015, 16:26
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In the rectangular coordinate system, the vertices of a triangle have coordinates (-3,0), (3,2), and (0,11). What is the area of the triangle.

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4.30
5.36
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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 02 Mar 2015, 00:03
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Turkish wrote:
In the rectangular coordinate system, the vertices of a triangle have coordinates (-3,0), (3,2), and (0,11). What is the area of the triangle.

1.15
2.18
3.24
4.30
5.36


hi turkish,
you can find the length of three sides..
1)\(\sqrt{( (-3-3)^2+(0-2)^2)}\)=\(\sqrt{40}\) say side A
2)\(\sqrt{( (-3-0)^2+(0-11)^2)}\)=\(\sqrt{130}\) say side B
3)\(\sqrt{( (3-0)^2+(2-11)^2)}\)=\(\sqrt{90}\).. say side C

now you can see A^2 + c^2 = B^2... this means there is a right angle between sides A and C..
so area=1/2*A*C=1/2*\(\sqrt{40}\)*\(\sqrt{90}\)=30..
ans D...
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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 02 Mar 2015, 00:12
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Hi Turkish,

I'm going to offer some suggestions so that you can physically attempt this question on your own:

1) Draw a quick graph/sketch and plot those 3 co-ordinates. Connect the co-ordinates with lines. The graph doesn't have to be perfect.
2) Draw a rectangle "around" this triangle. Your rectangle should have the following 4 corners:
(-3,0), (3,0), (-3,11), (3,11)

3) You should see your original triangle and 3 right triangles.
4) Figure out the area of the rectangle and subtract the areas of the 3 right triangles. This will give you the area of the original triangle.

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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 15 Jan 2016, 13:51
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Approximate by supposing that the vertex (3,2) is on the x-axis forming an isosceles triangle:
(3,2) -> (something>3;0)

Then, the area of this approximate isosceles triangle would be 6*11/2 = 33
The area of the actual triangle must be a little bit less than that of the approximate triangle. Thus 30, answer D.
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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 23 Apr 2017, 13:32
Here is an alternate and a bit easier approach provide you know how to solve a 3x3 matrix => USE THE DETERMINANT METHOD.
Area of a triangle => 1/2|Determinant Value|


Lets get the determinant of the matrix =>

1 -3 0
1 3 2
1 0 11

Determinant => 1(33) - (-3)(11-2) + 0(0-3) => 33+27=60

Hence the area of the triangle => 1/2 * 60 => 30



SMASH THAT D.

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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 26 Apr 2017, 01:40
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area of triangle = ½ |y₁ (x₂ - x₃) + y₂ (x₃ - x₁) + y₃ (x₁ - x₂)|
=1/2 { 27 + 33}
=30
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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 15 May 2017, 13:22
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Using Matrix Method
|-3 0 1|
| 3 2 1|
| 0 11 1|

-3(2-11)+33=60
taking half of matrix=30
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Re: In the rectangular coordinate system, the vertices of a triangle have [#permalink]

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New post 16 May 2017, 00:51
1/2 ( -3(2-11)+3(11-0)+0(0-2)) = 30
Re: In the rectangular coordinate system, the vertices of a triangle have   [#permalink] 16 May 2017, 00:51
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In the rectangular coordinate system, the vertices of a triangle have

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