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The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

The correct answer is D.

How do we know that the area is square and not something else ? _________________

The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

The correct answer is D.

How do we know that the area is square and not something else ?

Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0):

Attachment:

MSP254019e3df2hii74417g00000g7c2817hhbe0a3b.gif [ 2.86 KiB | Viewed 2271 times ]

Now, diagonals of the rectangle are equal (10 and 10), and also are perpendicular bisectors of each other (as the are on X and Y axises), so it must be a square. Area of a square equals to diagonal^2/2=10^2/2=50.

Re: M06#5 Absolute value [#permalink]
25 Jan 2011, 09:01

1

This post received KUDOS

Expert's post

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:

Attachment:

MSP86219e41c890f2h195700001928g4634f4e6h4h.gif [ 1.79 KiB | Viewed 2233 times ]

|x|+|y| =5:

Attachment:

MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 2231 times ]

Re: M06#5 Absolute value [#permalink]
26 Feb 2011, 16:42

1

This post received KUDOS

Expert's post

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.

Follow the links in my previous posts for more. _________________

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

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

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3 x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?

Merging similar topics. Please ask if anything remains unclear.

Yes, why can't it be a circle?

Well it cannot be a circle because it's a square. The reason why it's a square is given in previous posts (you can also see a diagram there).

Next, in an xy plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\). There is no resemblance whatsoever between this equation and the equation given in the question.

Re: M06#5 Absolute value [#permalink]
14 Apr 2014, 21:04

Hi Bunuel,

I understood your concept of equations and plotting points. But, i have a small doubt.

Why can't the graph in the second question (|x|+|y| =5 )be a horizontal square than a square turned by 45 degrees in the first question (|x+y|+|x-y|=4)

My question is there any reason why in the second question we have plotted a graph which is turned by 45 degrees rather than a horizontal square.

Help is appreciated.

Thanks in advance

Bunuel wrote:

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:

Attachment:

MSP86219e41c890f2h195700001928g4634f4e6h4h.gif

|x|+|y| =5:

Attachment:

MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif

gmatclubot

Re: M06#5 Absolute value
[#permalink]
14 Apr 2014, 21:04