The area bounded by the three curves |x + y | = 1, |x| = 1, and |y| =

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!

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.

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....

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

Bunuel - We get x+y= -1 and x+y= 1. How do I plot the given equation |x + y | = 1 on the graph?

area = 6*0.5*1*1 = 3

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

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.

The region is the unshaded portion in middle. The area has to be enclosed by all the three lines.

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.

chetan2u
Okay. Thanks a lot.

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