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# Equation |x| + |y| = 5 encloses a certain region on the

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Director
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Equation |x| + |y| = 5 encloses a certain region on the [#permalink]

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17 Feb 2007, 14:04
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Equation |x| + |y| = 5 encloses a certain region on the coordinate plane. What is the area of this region?

5
10
25
50
100
SVP
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17 Feb 2007, 15:07
(D) for me

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50
Director
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17 Feb 2007, 15:13
Fig wrote:
(D) for me

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50 :)
Fig, I don't follow. Please explain the bolded portion.
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17 Feb 2007, 15:20
ggarr wrote:
Fig wrote:
(D) for me

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50 :)
Fig, I don't follow. Please explain the bolded portion.

To know the area, I build up it by using the area of 4 similar triangles .... Then, I calculate the area of one of them. This area is also the area of the 3 other triangles.

Have a look on the Fig 1, the triangle in flashy green is the one described in my calculation of the area.
Attachments

Fig1_1 Triangle_The square.gif [ 4.85 KiB | Viewed 1391 times ]

Manager
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18 Feb 2007, 00:36
I another way you can go if you want to stick to squares is 5sqrt(2) x 5sqrt(2).
Director
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18 Feb 2007, 00:58
ok Fig,

I got ya'. but could we do the same with 1,4; -1,-4; -1,4; 1,-4 or 3,2;
-3,-2; -3,2; 3,-2. I tried them cursorily. they didn't seem to work.
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18 Feb 2007, 01:25
ggarr wrote:
ok Fig,

I got ya'. but could we do the same with 1,4; -1,-4; -1,4; 1,-4 or 3,2;
-3,-2; -3,2; 3,-2. I tried them cursorily. they didn't seem to work.

The 2 shapes u design like this are rectangles. ... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|) .... If they were equal, then we have a square

Notice also that sides of such rectangle are parallel to the X and Y axes

If gives a and b the absolute values of the x and y coordonaites of each vertex from a same rectangle, then we have:

The aera of such a rectangle = (2*a) * (2*b)

o Aera 1 = 2*1 * 2*4 = 16
o Aera 2 = 3*1 * 2*2 = 12
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18 Feb 2007, 02:52
|x| + |y| = 5
Quote:
The 2 shapes u design like this are rectangles. ... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|
In the stem, we need the abs. val. of x and y to equal 5. Why would we need the abs. val. of x and y (|1| and |4| in this case) to be equal. Don't they just need to sum to 5?
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18 Feb 2007, 03:06
ggarr wrote:
|x| + |y| = 5
Quote:
The 2 shapes u design like this are rectangles. ... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|
In the stem, we need the abs. val. of x and y to equal 5. Why would we need the abs. val. of x and y (|1| and |4| in this case) to be equal. Don't they just need to sum to 5?

No .... It is 2 different things.... Your example are not equivalent to the problem

|x| + |y| = b (b>0) will always create a square such that it is shown on my Fig 1.

Your examples create rectangles with sides that are parellel to the X and Y axes .... Draw them, u will see what I mean

An equivalent system of equations for your examples is :
o |x| = c
o |y| = d
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Re: Coordinate Geometry problem 2 [#permalink]

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21 Aug 2009, 17:26
Four different approaches to reach to a quick answer:

1) Diagonal of the square. d=10 -> area = 10^2 / 2 = 50

2) Phytagoras, 45:45:90. Hyp=side=5*sqrt(2) -> area = (5*sqrt(2))^2 = 50

3) Phytagoras, calculation. Hyp=side=sqrt(5^2+5^2)=sqrt(50) -> area = (sqrt(50))^2 = 50

4) 4 triangles. areaT = (b*h)/2 = (5*5)/2 = 25/2 -> area = areaT*4 = 25/2 * 4 = 50
Re: Coordinate Geometry problem 2   [#permalink] 21 Aug 2009, 17:26
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