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If equation |x| + |y| = 5 encloses a certain region on the

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Manager
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If equation |x| + |y| = 5 encloses a certain region on the [#permalink]

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17 Jan 2008, 23:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If equation |x| + |y| = 5 encloses a certain region on the graph, what is the area of that region?

* 5
* 10
* 25
* 50
* 100

i don't even know how to evaluate a statement like that...
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17 Jan 2008, 23:46
1
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Expert's post
D

Just draw it.
we have a square with diagonal=5*2=10

$$Area=a^2=\frac{(\sqrt2*a)^2}{2}=\frac{(diagonal)^2}{2}=\frac{(10)^2}{2}=50$$
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Manager
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18 Jan 2008, 00:06
walker wrote:
D

Just draw it.
we have a square with diagonal=5*2=10

$$Area=a^2=\frac{(\sqrt2*a)^2}{2}=\frac{(diagonal)^2}{2}=\frac{(10)^2}{2}=50$$

sweet... kudos +1... check back on the exponent question i had for me.
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18 Jan 2008, 06:55
my approach was similar to walker's - it's important to draw it to visualize, then obviously the area of each triangle is 5*5/2 = 25/2 and there are 4 triangles. the answer is 50.
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19 Jan 2008, 01:41
walker wrote:
D

Just draw it.
we have a square with diagonal=5*2=10

$$Area=a^2=\frac{(\sqrt2*a)^2}{2}=\frac{(diagonal)^2}{2}=\frac{(10)^2}{2}=50$$

At first, I dont know how to solve this problem, but when I see your drawing, I read it as:

Keep x=0, you had y = + - 5, and vice versa, keep y=0, you had x = +-5.

Do you have any tips for absolute value problem? When problem involves x and y, I am in block.
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19 Jan 2008, 02:05
sondenso wrote:
Do you have any tips for absolute value problem? When problem involves x and y, I am in block.

The right but long approach is:

$$|x| + |y| = 5$$

is equal:

$$x+y=5\text{ if x\ge0;y\ge0}$$
$$x-y=5\text{ if x\ge0;y<0}$$
$$-x+y=5\text{ if x<0;y\ge0}$$
$$-x-y=5\text{ if x<0;y<0}$$

shortcut:

your drawing have to be symmetric for x and y. Therefore you can draw only x+y=5 (I quadrant) and use mirror symmetry
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07 Feb 2008, 00:19
can we take 3,2 or 4,1 as points?
Since the absolute value will always be "5".

-Jack
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07 Feb 2008, 00:54
jackychamp wrote:
can we take 3,2 or 4,1 as points?
Since the absolute value will always be "5".

-Jack

we can but these points are not vertexs of the square and therefore, the points cannot help us to find area.
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Re: absolute value problem   [#permalink] 07 Feb 2008, 00:54
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