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two sides lie

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by sanju09 » Tue Oct 12, 2010 10:53 pm
What is the area of the square whose two sides lie on the lines x + y = 2 and x + y = -2?
(A) 4 √2
(B) 8
(C) 8 √2
(D) 16
(E) 16 √2



[spoiler]Made up[/spoiler]
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by shovan85 » Wed Oct 13, 2010 1:23 am
sanju09 wrote:What is the area of the square whose two sides lie on the lines x + y = 2 and x + y = -2?
(A) 4 √2
(B) 8
(C) 8 √2
(D) 16
(E) 16 √2



[spoiler]Made up[/spoiler]
IMO B.

From the eqn x+y=2 and x+y=-2 we can get the points where the lines intersect X and Y axes.
The 4 points will be (2,0), (0,2), (-2,0), (0,-2).

These 4 points are the vertices as the X and Y axis distance between these points are 4 units each and they are perpendicular to each other.

Thus any one side can be the distance between (2,0) and (0,2) = 2*sqrt(2)

Thus area [2*sqrt(2)]^2 = 8 .
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by neerajkumar1_1 » Wed Oct 13, 2010 1:23 am
the two lines are parallel with slope -1...
all we need to do is get the distance between the two lines... which will give the area of any square formed between the two lines...

there are several methods to find the distance between the two lines...
i have used the following..

y = x will be a line perpendicular to both the lines with a slope of 1

solving both the given lines with y=x
will give the points 1,1 and -1,-1 on the respective lines...

the distance between the two will = 2 root(2)* 2 root (2)
thus the area will be 8

Pick B
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by Arcane66 » Wed Oct 13, 2010 9:16 am
I also got B but I wasn't really sure. I graphed both of the lines and then simply drew a vertical line cutting what I thought the square would be in half. I got 4 for the diagonal of the square. Then you just use the x:x:xsqrt2 formula and find the sides of the square equal 2sqrt2. Squaring 2sqrt2 gives you 8. So I think B? You should work for one of these companies, haha. This took more than 2 minutes :(.
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by shovan85 » Wed Oct 13, 2010 9:57 am
Arcane66 wrote: I got 4 for the diagonal of the square. Then you just use the x:x:xsqrt2 formula and find the sides of the square equal 2sqrt2. Squaring 2sqrt2 gives you 8. So I think B? You should work for one of these companies, haha. This took more than 2 minutes :(.
FYI: After getting diagonal you can use (D^2)/2 [D=diagonal of square]
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by bdevas01 » Fri Oct 15, 2010 5:34 am
I'm not sure what I'm doing wrong here..4 coordinates are (0,2) (2,0) (0, -2) and (-2, 0).

That means that the diagonal of the square is 4 so each side is 4/(sqrt 2).

Area = base x height = 4/(sqrt 2) * 4/(sqrt 2) = 8 x 1/2 = 4.

But thats not one of the answer..
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by sanju09 » Fri Oct 15, 2010 6:14 am
bdevas01 wrote:I'm not sure what I'm doing wrong here..4 coordinates are (0,2) (2,0) (0, -2) and (-2, 0).

That means that the diagonal of the square is 4 so each side is 4/(sqrt 2).

Area = base x height = 4/(sqrt 2) * 4/(sqrt 2) = 8 x 1/2 = 4.

But thats not one of the answer..

...because 4/(sqrt 2) * 4/(sqrt 2) = 16 x 1/2 = 8, man in hurry
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by bdevas01 » Fri Oct 15, 2010 6:25 am
Sanju, i am not sure what you are implying.

The area of a triangle is (1/2)*(b*h)

Base X Height is 4/(sqrt 2) * 4/(sqrt 2), which I am certain is equal to 16/2 which equals 8.

Now 1/2 * 8 (because we still need to multiply by 1/2) = 4.

But 4 isn't an answer choice...
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by shovan85 » Fri Oct 15, 2010 7:08 am
bdevas01 wrote:Sanju, i am not sure what you are implying.

The area of a triangle is (1/2)*(b*h)

Base X Height is 4/(sqrt 2) * 4/(sqrt 2), which I am certain is equal to 16/2 which equals 8.

Now 1/2 * 8 (because we still need to multiply by 1/2) = 4.

But 4 isn't an answer choice...
Area of the Square is being asked... why r u bothering about the Triangle ;)
Read the question properly my friend...
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by bdevas01 » Fri Oct 15, 2010 7:26 am
shovan85 wrote:
bdevas01 wrote:Sanju, i am not sure what you are implying.

The area of a triangle is (1/2)*(b*h)

Base X Height is 4/(sqrt 2) * 4/(sqrt 2), which I am certain is equal to 16/2 which equals 8.

Now 1/2 * 8 (because we still need to multiply by 1/2) = 4.

But 4 isn't an answer choice...
Area of the Square is being asked... why r u bothering about the Triangle ;)
Read the question properly my friend...
Got it! Thanks, I know such little errors in reading are common on the GMAT. I'll be taking things a slightly slower pace from now on.

Thanks again guys!
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by pesfunk » Mon Nov 01, 2010 4:59 pm
I drew the 2 lines and found 2 points 1,1 and -1,-1

Now, assuming that one of the sides of the square is the line touching the above 2 points,
I found the distance between these 2 points

distance = root [ (x1-x2)^2 + (y1-y2)^2) ]

found one of the sides as root (8)

So the area should be 8.
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