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sunny3011
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I am sorry but I don't understand how do we arrive at the shaded region whose area is required. So, can anyone out there help me out? Thanks.

Regards,
Pallavi
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pallaviisinha
I am sorry but I don't understand how do we arrive at the shaded region whose area is required. So, can anyone out there help me out? Thanks.

Regards,
Pallavi

We have to make a graph with 6 equations:
x=1
x=-1
y=1
y=-1
x+y=1
x+y=-1

The shaded region is one with the overlap..two squares of side 1 & two right triangles with area 1/2 each..which gives the answer as 3.

Unfortunately in the photo accompanying the explanation, in place of the line x=-1, the line x=-2 is drawn. Perhaps that caused the confusion.

Hope it helped!
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ytsejam
pallaviisinha
I am sorry but I don't understand how do we arrive at the shaded region whose area is required. So, can anyone out there help me out? Thanks.

Regards,
Pallavi

We have to make a graph with 6 equations:
x=1
x=-1
y=1
y=-1
x+y=1
x+y=-1

The shaded region is one with the overlap..two squares of side 1 & two right triangles with area 1/2 each..which gives the answer as 3.

Unfortunately in the photo accompanying the explanation, in place of the line x=-1, the line x=-2 is drawn. Perhaps that caused the confusion.

Hope it helped![/quoteTh

Indeed, thanks.
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sunny3011
The area bounded by the three curves |x + y | = 1, | x | = 1, and | y | = 1, is equal to
a. 1
b. 2
c. 3
d. 4
e. 8


I have tried to plot graph of lines and I got area from |x|=1 and |y|=1. but what about line |x+y|=1???
Confused and need to clarify the concepts?

Source: GMAT Tutor Question bank


|x+y|=1 can be written as x+y=1 or x+y=-1. These are two straight lines, which are mirror image of each as shown in the following graph.

|x|=1 and |y|=1 are also drawn on the graph with red and blue lines respectively. The required area is shown in the grey blocks. which is equal to 3(2+1/2+1/2). (2 block of area 1 and two triangles of area =1/2)

The solution is correct.
However,I think the line x=-1 is not represented properly here...
It should just be shifted to right side so that the area is covered by all the lines...
2 square boxes and two triangles must be covered by all the lines....
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manpreetsingh86
sunny3011
The area bounded by the three curves |x + y | = 1, | x | = 1, and | y | = 1, is equal to
a. 1
b. 2
c. 3
d. 4
e. 8


I have tried to plot graph of lines and I got area from |x|=1 and |y|=1. but what about line |x+y|=1???
Confused and need to clarify the concepts?

Source: GMAT Tutor Question bank


|x+y|=1 can be written as x+y=1 or x+y=-1. These are two straight lines, which are mirror image of each as shown in the following graph.

|x|=1 and |y|=1 are also drawn on the graph with red and blue lines respectively. The required area is shown in the grey blocks. which is equal to 3(2+1/2+1/2). (2 block of area 1 and two triangles of area =1/2)

The red line should be 1 square to the right? looks like it represents x=-2 instead of x=-1
Please correct me if Im wrong
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Bunuel - We get x+y= -1 and x+y= 1. How do I plot the given equation |x + y | = 1 on the graph?
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area = 6*0.5*1*1 = 3

Bunuel nick1816 VeritasKarishma
are the shaded regions bounded by these curves correct?
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region.jpg [ 1.48 MiB | Viewed 13577 times ]

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sunny3011
The area bounded by the three curves |x + y | = 1, |x| = 1, and |y| = 1, is equal to

A. 1
B. 2
C. 3
D. 4
E. 8


I have tried to plot graph of lines and I got area from |x|=1 and |y|=1. but what about line |x+y|=1???
Confused and need to clarify the concepts?

Source: GMAT Tutor Question bank

See the attached figure.

Now |x|=1 will give two straight lines perpendicular to x-axis crossing x at 1 and the other at -1.
Similarly |y|=1 will give two straight lines perpendicular to y-axis crossing y at 1 and the other at -1.

Thus |x|=1 and |y|=1 enclose a square of side 1-(-1)=2, so area of square = 2*2=4. (also shown by blue colour in graph).
So our area has to be less than or at the max equal to 4
. Eliminate E.

Draw lines |x+y|=1. The area which it excludes from 4 is shown by the star.
Now each square in the graph is 1*1, and |x|=1 and |y|=1 includes 4 such squares of area 1 each.=>184=1
But each of the \(\triangle\) shown by the two stars are equal to half of the square 1*1. Thus, area of two triangles combined is equal to that of one square of 1*1 or 1.
Area, therefore, covered by |x|=1, |y|=1, and |x+y|=1 is 4-1=3.

Also since each of the |x|=1, |y|=1, and |x+y|=1 are linear equations and therefore equation of lines, it would be appropriate to mention them as lines rather than curve.
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preetamsaha
area = 6*0.5*1*1 = 3

Bunuel nick1816 VeritasKarishma
are the shaded regions bounded by these curves correct?


The region is the unshaded portion in middle. The area has to be enclosed by all the three lines.
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preetamsaha
area = 6*0.5*1*1 = 3

Bunuel nick1816 VeritasKarishma
are the shaded regions bounded by these curves correct?

chetan2u
in my picture, all the areas are enclosed by three lines. am i wrong? although in your approach I found it very much logical.
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preetamsaha
preetamsaha
area = 6*0.5*1*1 = 3

Bunuel nick1816 VeritasKarishma
are the shaded regions bounded by these curves correct?

chetan2u
in my picture, all the areas are enclosed by three lines. am i wrong? although in your approach I found it very much logical.


No, when we say area enclosed by |x|=1, all the area above the line x=1 and below the area x=-1 will be discarded.
So all the ares you have chosen is actually out of at least one of the regions enclosed by |x|=1, or by |y|=1 or by |x+y|=1.

Most of the time, when you talk of 'enclosed by lines' that are on either side of ORIGIN, the area that includes the origin is the correct portion.
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chetan2u
Okay. Thanks a lot.
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