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Equation |x/2| + |y/2| = 5 encloses a certain region on the

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Director
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Equation |x/2| + |y/2| = 5 encloses a certain region on the [#permalink] New post 20 Dec 2005, 04:16
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A
B
C
D
E

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Equation |x/2| + |y/2| = 5 encloses a certain region on the coordiante plane. What is the area of the region

A 20
B 50
C 100
D 200
E 400

The line of argument I use is insufficient, since it leads always to 100. Please give me an explanation other than that provided in the challenge. Thanks
Director
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 [#permalink] New post 20 Dec 2005, 05:23
Equation |x/2| + |y/2| = 5 encloses a certain region on the coordiante plane. What is the area of the region?

1) If x < 0 and y < 0, -(x/2)-(y/2)=5 ---> y = -x - 10
2) If x < 0 and y > 0, -(x/2)+(y/2)=5 ---> y = x + 10
3) If x > 0 and y < 0, (x/2)-(y/2)=5 ---> y = x - 10
4) If x > 0 and y > 0, (x/2)+(y/2)=5 ---> y = -x + 10

When we draw a graph using the above 4 equtations, we will have a square one of whose length is 10root2.

10root2 * 10root2 = 200

I vote for (D) 200.
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 [#permalink] New post 20 Dec 2005, 06:42
my edited message..

the side of the square is 10sqrt2, and hence the area is 200.

When x=0, Y is +/- 10
similarly when Y=0, x is +/-10


Explanation:

if these points are drawn on a x/y plane it make a square, with the side

sqrt (10*10) +(10*10)= 10sqrt2

and hence the area is 200.

I hope it is clear TO YOU ---- ALLABOUT

Last edited by SunShine on 20 Dec 2005, 12:38, edited 3 times in total.
Director
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 [#permalink] New post 20 Dec 2005, 07:34
gamjatang wrote:
Equation |x/2| + |y/2| = 5 encloses a certain region on the coordiante plane. What is the area of the region?

1) If x < 0 and y < 0, -(x/2)-(y/2)=5 ---> y = -x - 10
2) If x < 0 and y > 0, -(x/2)+(y/2)=5 ---> y = x + 10
3) If x > 0 and y < 0, (x/2)-(y/2)=5 ---> y = x - 10
4) If x > 0 and y > 0, (x/2)+(y/2)=5 ---> y = -x + 10

When we draw a graph using the above 4 equtations, we will have a square one of whose length is 10root2.

10root2 * 10root2 = 200

I vote for (D) 200.


The first step is clear, but unfortunately the second step not. We have four ( three) equations, I see them infront of me, but why should the sides be 10sqrt2?

The OA is D
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 [#permalink] New post 20 Dec 2005, 16:09
allabout wrote:
gamjatang wrote:
Equation |x/2| + |y/2| = 5 encloses a certain region on the coordiante plane. What is the area of the region?

1) If x < 0 and y < 0, -(x/2)-(y/2)=5 ---> y = -x - 10
2) If x < 0 and y > 0, -(x/2)+(y/2)=5 ---> y = x + 10
3) If x > 0 and y < 0, (x/2)-(y/2)=5 ---> y = x - 10
4) If x > 0 and y > 0, (x/2)+(y/2)=5 ---> y = -x + 10

When we draw a graph using the above 4 equtations, we will have a square one of whose length is 10root2.

10root2 * 10root2 = 200

I vote for (D) 200.
The first step is clear, but unfortunately the second step not. We have four ( three) equations, I see them infront of me, but why should the sides be 10sqrt2?



It's the Pythagorean theorem. ( a^2 + b^2 = c^2 )

Or you can just simply think that there are 4 triangles.

The OA is D

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 [#permalink] New post 20 Dec 2005, 18:00
gamjatang,

Excellent job. This is a very good question to learn and apply several topics.
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 [#permalink] New post 21 Dec 2005, 02:00
Thanks all.
Don't know why I did not see the nulls.
  [#permalink] 21 Dec 2005, 02:00
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Equation |x/2| + |y/2| = 5 encloses a certain region on the

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