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dheeraj24
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In the x-y plane, the area of the region bounded by the graph of |x + 14 Apr 2014, 21:48
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Hi karishma,

Could you please elaborate more on this one .
I am really having a hard time figuring this one out .
I understood the first question i.e
In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16.
But in Second Question
what is the area bounded by graph|x/2| + |y/2| = 5?
Why was the graph drawn different from that of the previous Question.
Why was diagonal considered for second question and Side of a square considered for first question.

Help is appreciated .
Thanks in advance :)
VeritasPrepKarishma
pawankumargadiya
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8

The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect.
You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise.
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In the x-y plane, the area of the region bounded by the graph of |x + 14 Apr 2014, 22:43
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dheeraj24
Hi karishma,

Could you please elaborate more on this one .
I am really having a hard time figuring this one out .
I understood the first question i.e
In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16.
But in Second Question
what is the area bounded by graph|x/2| + |y/2| = 5?
Why was the graph drawn different from that of the previous Question.
Why was diagonal considered for second question and Side of a square considered for first question.

Help is appreciated .
Thanks in advance :)
Show more

The graph was drawn differently because the equations given to you are different.

|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?

y = 10+x
y = 10 -x
y = x - 10
y = -x - 10

When you draw them out, you get slanting lines and the figure which looks like a kite.
The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.

In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side.
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In the x-y plane, the area of the region bounded by the graph of |x + 14 Apr 2014, 23:18
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Yeah karishma,

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

Thanks in advance.

VeritasPrepKarishma
dheeraj24
Hi karishma,


The graph was drawn differently because the equations given to you are different.

|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?

y = 10+x
y = 10 -x
y = x - 10
y = -x - 10

When you draw them out, you get slanting lines and the figure which looks like a kite.
The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.

In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side.
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In the x-y plane, the area of the region bounded by the graph of |x + 15 Apr 2014, 05:01
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dheeraj24
Yeah karishma,

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

Thanks in advance.

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Because you are asked the area of the region bounded by |x+y| + |x-y| = 4.
This equation gives you ONLY horizontal/vertical lines passing through points (2,0), (-2,0), (0,2) and (0,-2) such as x = 2, y = 2, x = -2, y = -2.

Note that x = 2 is the equation of a line (it is not a coordinate) which passes through point (2, 0) and is parallel to the y axis. Similarly, y = 2 is the equation of a line which is parallel to x axis and passes through the point (0, 2) and so on. I think you are taking x = 2 as a coordinate but that is not the case. A coordinate has a value for y too. x =2 is the equation of a line. It implies that x coordinate is always 2 and y can be anything. So all points lying on a line passing through x = 2 and parallel to y axis satisfy this criteria.
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In the x-y plane, the area of the region bounded by the graph of |x + 15 Apr 2014, 09:02
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Hi Karishma,

I am still not clear :(

in question |x/2| + |y/2| = 5 we are getting following cordinates.

x=10
x=-10
y=10
y=-10

and in question |x+y| + |x-y| = 4. we are having following cordinates
x=2
x=-2
y=2
y=-2

why we are drawing graph differently?

Thanks
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In the x-y plane, the area of the region bounded by the graph of |x + 15 Apr 2014, 09:16
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Hi Karishma,

I am still not clear :(

in question |x/2| + |y/2| = 5 we are getting following cordinates.

x=10
x=-10
y=10
y=-10

and in question |x+y| + |x-y| = 4. we are having following cordinates
x=2
x=-2
y=2
y=-2

why we are drawing graph differently?

Thanks
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For \(|x+y| + |x-y| = 4\) the equations are:

\(x = 2\);
\(y = 2\);
\(y = -2\);
\(x = -2\).

For \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) the equations are:

\(y=-10-x\);
\(y=10+x\);
\(y=10-x\);
\(y=x-10\).

You might helpful to re-read the thread.
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In the x-y plane, the area of the region bounded by the graph of |x + 20 Jun 2015, 02:58
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Bunuel
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Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph\(|x/2| + |y/2| = 5\)?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

After solving you'll get equation of four lines:

\(y=-10-x\)
\(y=10+x\)
\(y=10-x\)
\(y=x-10\)

These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).


Hope it's clear.
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Hi bunnel,

How did u rhombus for this one and a square for the other one?...I got the limits for both the questions, but could not figure out they turn out to be a square and rhombus!...
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jayanthjanardhan
Bunuel
srini123
Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph\(|x/2| + |y/2| = 5\)?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

After solving you'll get equation of four lines:

\(y=-10-x\)
\(y=10+x\)
\(y=10-x\)
\(y=x-10\)

These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).


Hope it's clear.

Hi bunnel,

How did u rhombus for this one and a square for the other one?...I got the limits for both the questions, but could not figure out they turn out to be a square and rhombus!...
Show more

Even that is a square but never forget that A Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that figure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won't be wrong either but you are right about the figure being a Square.
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In the x-y plane, the area of the region bounded by the graph of |x + 14 Oct 2024, 02:42
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`Hey I did something rather stupid and I need help in identifying what is exact mistake

|x+y|+|x-y| = 4
Why cant we do |x+y|=A and |x-y|=B

So now a few values of A & B could be,
A,B =
0,4
4,0
2,2

Using First
A=0 and B=4
|x+y|=0 Hence (x,y) could be (0,0)
|x-y|=4 Hence (x,y) could be (0,4),(4,0)
And so on.

Bunuel just help me understand what am I doing wrong here even though I agree (4,0) and (0,4) cant be the solution. Just need to know so I dont make this mistake again
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In the x-y plane, the area of the region bounded by the graph of |x + 14 Oct 2024, 02:52
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MalachiKeti
`Hey I did something rather stupid and I need help in identifying what is exact mistake

|x+y|+|x-y| = 4
Why cant we do |x+y|=A and |x-y|=B

So now a few values of A & B could be,
A,B =
0,4
4,0
2,2

Using First
A=0 and B=4
|x+y|=0 Hence (x,y) could be (0,0)
|x-y|=4 Hence (x,y) could be (0,4),(4,0)
And so on.

Bunuel just help me understand what am I doing wrong here even though I agree (4,0) and (0,4) cant be the solution. Just need to know so I dont make this mistake again
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­I've tried explaining this question in several posts on the previous two pages. However, if it's still unclear, you can ignore this question since geometry is no longer part of the GMAT syllabus.
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In the x-y plane, the area of the region bounded by the graph of |x + 17 Jul 2025, 21:08
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Hi Brunel,

I think it still looks the same with the |x/2|+|y/2|=5 problem.

Solving this we can get

x=10
y=10
x=-10
y=-10

Just like this problem.

I get that you used two different methods to solve these two. But based on what one will understand which method to follow?
What is the actual difference between these two questions?

Bunuel
srini123
Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph\(|x/2| + |y/2| = 5\)?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

After solving you'll get equation of four lines:

\(y=-10-x\)
\(y=10+x\)
\(y=10-x\)
\(y=x-10\)

These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).


Hope it's clear.
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In the x-y plane, the area of the region bounded by the graph of |x + 17 Jul 2025, 21:19
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Hi Brunel,

Didn't understand the formula with area and diagonal wen solving the |x/2| + |y/2| = 5 problem. Is area of a rectangle is equal to square of the diagonal divided by 2?

Bunuel
srini123
Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph\(|x/2| + |y/2| = 5\)?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

After solving you'll get equation of four lines:

\(y=-10-x\)
\(y=10+x\)
\(y=10-x\)
\(y=x-10\)

These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).


Hope it's clear.
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Bunuel
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­I've tried explaining this question in several posts on the previous two pages. However, if it's still unclear, you can ignore this question since geometry is no longer part of the GMAT syllabus.
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