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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 2334 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 2296 times ]
|x|+|y| =5:
Attachment:
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 2294 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