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New post 16 Sep 2014, 00:27
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New post 16 Sep 2014, 00:27
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Official Solution:


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


A. 5
B. 10
C. 25
D. 50
E. 100


X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

Image

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to \(\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50\).


Answer: D
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New post 19 Oct 2015, 12:46
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Bunuel wrote:
Official Solution:


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


A. 5
B. 10
C. 25
D. 50
E. 100


X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

Image

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to \(\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50\).


Answer: D



Another way to find the area of the square, if you don't know how to find it from the diagonal, is to take the area of the four triangles that are made from the x and y axis. You end up with \((\frac{5*5}{2})*4\).
=
Additionally, you can find the difference between any two points with the formula \(\sqrt{(X2-X1)^2+(Y2-Y1)^2}\). Plug in (5,0) and (0,5) into that equation, then square the result to find the area.
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New post 20 Jul 2016, 08:06
Thanks B

one more question please

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.
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New post 30 Jul 2016, 12:14
shrive555 wrote:
Thanks B

one more question please

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.


You can find the area of a square in either way: \(area_{square}=side^2=\frac{diagonal^2}{2}\).

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees.
|x+y|+|x-y|=4:
Image

|x|+|y| =5:
Image


Attachment:
MSP86219e41c890f2h195700001928g4634f4e6h4h.gif

Attachment:
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif

>> !!!

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New post 30 Jul 2016, 12:35
Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong
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New post 03 Aug 2016, 08:17
ajit257 wrote:
Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong


We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure.

If x>0 and y>0 we'll have x+y=5;
If x>0 and y<0 we'll have x-y=5;
If x<0 and y>0 we'll have -x+y=5;
If x<0 and y<0 we'll have -x-y=5;

So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them.
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New post 22 Aug 2016, 04:39
I think this is a high-quality question and I agree with explanation.
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New post 19 Jun 2017, 18:21
is there a way to know the sides from the information given? thx
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New post 22 Sep 2017, 04:04
I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say :-)

TIA!
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New post 22 Sep 2017, 04:38
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susheelh wrote:
I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say :-)

TIA!


|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

Hope it's clear.
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New post 22 Sep 2017, 04:42
A million thanks to you Bunuel for the quick answer. I fully understand this now. Awesome solution to an equally awesome question!

Bunuel wrote:
susheelh wrote:
I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say :-)

TIA!


|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

Hope it's clear.

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New post 29 May 2018, 03:16
Hi,
I got the points as (5,5), (5,-5), (-5,5) and (-5,-5). This makes each side 10 units and is a square. So the area is 100. Where am I going wrong? I got the points by substituting x=0 and y=0 for each equation of the line |x| + |y| = 5.
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New post 29 May 2018, 07:39
dishadesai wrote:
Hi,
I got the points as (5,5), (5,-5), (-5,5) and (-5,-5). This makes each side 10 units and is a square. So the area is 100. Where am I going wrong? I got the points by substituting x=0 and y=0 for each equation of the line |x| + |y| = 5.


How did you get those values if you substituted x = 0 and y = 0?

|x| + |y| = 5:

If x = 0, then y is 5 or -5 --> two y-intercepts (0, 5), and (0, -5).
If y = 0, then x is 5 or -5 --> two x-intercepts (5, 0), and (-5, 0).

So, X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0)
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New post 29 Jul 2018, 09:00
Bunuel niks18 chetan2u generis pushpitkc KarishmaB EMPOWERgmatRichC GMATPrepNow

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


It is interesting to see jump from absolute modulus to intercepts in this problem.
Do we get a hint about this approach from highlighted text ?

I believe it is very difficult to solve |x| + |y| = 5 on its own. Let me know your thoughts.
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New post 30 Jul 2018, 04:21
adkikani wrote:
Bunuel niks18 chetan2u generis pushpitkc KarishmaB EMPOWERgmatRichC GMATPrepNow

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


It is interesting to see jump from absolute modulus to intercepts in this problem.
Do we get a hint about this approach from highlighted text ?

I believe it is very difficult to solve |x| + |y| = 5 on its own. Let me know your thoughts.


It is a co-ordinate geometry question (keywords are graph, area of the region). Just that the equations of the lines are given in a more compact format involving absolute values.
If you know your absolute values, you know the equations you will get are x + y = 5, x - y = 5 etc depending on the signs of x and y.
Also, it is very easy to visualise these lines by putting each variable equal to 0 in turn.
x + y = 5
If x = 0, y = 5 and if y = 0, x = 5 so line will intersect the axis at (0, 5) and (5, 0) and so on.
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New post 10 Jul 2019, 21:26
If there was an answer choice \(\frac{25}{2}\). I would have selected that without taking double look.
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