If equation |x/2|+|y/2| = 5 encloses a certain region

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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]

15 Jan 2012, 09:42

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If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400

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Re: Area of region [#permalink]

15 Jan 2012, 11:07

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Apex231 wrote:

I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?

Given: |x|+|y|=20

You will have 4 case:

x<0 and y<0 --> -x-y=10 --> y=-10-x;

x<0 and y\geq{0} --> -x+y=10 --> y=10+x;

x\geq{0} and y<0 --> x-y=10 --> y=x-10;

x\geq{0} and y\geq{0} --> x+y=10 --> y=10-x;

So we have equations of 4 lines. If you draw these four lines you'll get the figure shown in my previous post.

Hope it's clear.
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Re: Area of region [#permalink]

15 Jan 2012, 10:26

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Apex231 wrote:

Please move this to PS thread...

If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A 20
B 50
C 100
D 200
E 400

First of all to simplify the given expression a little bit let's multiply it be 2: |\frac{x}{2}|+|\frac{y}{2}|=5 --> |x|+|y|=10.

Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0):
y=0 --> |x|=10 --> x=10 and x=-10;
x=0 --> |y|=10 --> y=10 and y=-10.

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by |x|+|y|=10:

Attachment:

Enclosed region.gif [ 2.04 KiB | Viewed 2318 times ]

You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200.

Or the Side= \sqrt{200} --> area=side^2=200.

Answer: D.

Similar questions:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html

Hope it's clear.
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Re: Area of region [#permalink]

15 Jan 2012, 10:58

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Re: Area of region [#permalink]

15 Jan 2012, 12:10

Thanks so much Bunuel...

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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]

10 Sep 2012, 12:25

CMcAboy wrote:

Can someone help me with this question:

If equation |x/2| + |y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A) 20
B) 50
C) 100
D) 200
E) 400

I believe this is the simplest & the quickest solution
|x/2| + |y/2| = 5
Put x = 0 in the above equation we get |y/2| = 5, which means y= 10, - 10
Put y = 0 in the above equation we get |y/2| = 5, which means x= 10, - 10

If you see plot these four points you get a square with two equal diagonals of length 20 units
Thus area = 1/2 * (Diagonal)^2 -----> 1/2 * 400 = 200

I hope this will help many.
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]

05 Dec 2012, 23:30

(1) derive all equations
x+y = 10
x-y = 10
x+y=-10
-x+y=10

(2) Get your x and y intercepts

(0,10), (10,0)
(0,-10),(10,0)
(0,-10),(-10,0)
(0,10),(-10,0)

(3) You will have a square with a diagonal of 20
(4) Calculate area = (10 * \sqrt{2})\sqrt{^2} = 200

Answer : D

Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink] 05 Dec 2012, 23:30

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If equation |x/2|+|y/2| = 5 encloses a certain region

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If equation |x/2|+|y/2| = 5 encloses a certain region

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