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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 [#permalink] New post 15 Jan 2012, 08:42
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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] New post 15 Jan 2012, 10:07
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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] New post 15 Jan 2012, 09:26
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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
Enclosed region.gif [ 2.04 KiB | Viewed 5775 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

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Re: Area of region [#permalink] New post 15 Jan 2012, 09:58
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?
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Re: Area of region [#permalink] New post 15 Jan 2012, 11:10
Thanks so much Bunuel...
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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink] New post 10 Sep 2012, 11:25
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] New post 05 Dec 2012, 22:30
(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
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink] New post 12 Oct 2014, 02:26
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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink] New post 15 Oct 2014, 18:17
Apex231 wrote:
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



Hello There,
Equation of a straight line whose x and y intercepts are a and b resp. is (x/a) + (y/b) = 1 i.e., coordinates of two ends of the line are (a,0) and (0,b).
Now, from the given question,
|x/2|+|y/2| = 5, reducing this to intercept form we get,
|x/10|+|y/10| = 1
Considering the equation without modulus, coordinates are (10,0) and (0,10). Since there is modulus, other two coordinates are (-10,0) and (0,-10).
Now coordinates (10,0), (0,10), (-10,0) and (0,-10) form a square with diagonal length = 20.
Here diagonal length can be obtained by calculating the distance between (10,0) and (-10,0) or (0,10) and (0,-10).
In a square,
Diagonal = Side * sqrt(2)
Side = 10 * sqrt(2)
Area = Side * Side = 200.

Ans : D

Hope this helps!
Thanks!
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink] New post 15 Oct 2014, 23:48
Expert's post
Apex231 wrote:
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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Hope this helps.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If equation |x/2|+|y/2| = 5 encloses a certain region   [#permalink] 15 Oct 2014, 23:48
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