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To find the perimeter

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To find the perimeter

by gmattesttaker2 » Sat Feb 15, 2014 9:14 pm
Hello,

For the following:

Points A and C have (x,y) coordinates (-1,3) and (2,-1) respectively, and are
opposite vertices of square ABCD, what is the perimeter of square ABCD?


OA: [spoiler]10 sq. root (2)[/spoiler]


I was a bit confused by this question since it doesn't appear to be a square as shown in the diagram:

Thanks a lot,
Sri
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Quadrilateral ABCD
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by Patrick_GMATFix » Sat Feb 15, 2014 10:47 pm
Sri,

You assumed that the sides of the squares are parallel to the x and y axes. As you drawing shows, this is not possible since the shape you drew is not a square. Think of the question in more simple terms:

Given the length of a diagonal (the distance between two opposite vertices), what is the perimeter of the square? If you find the length of AC, you will have the first piece!
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by [email protected] » Sun Feb 16, 2014 12:41 am
Hi Sri,

As you deduced, and Patrick confirmed, the square that you're looking for is not parallel to the x-axis nor the y-axis. Be careful of the assumptions that you make on GMAT questions (this warning is most appropriate for DS questions, but applies to other questions as well - including this one) as an incorrect assumption could lead you to an incorrect answer (and at the very least, cost you time).

As it stands, you're not likely to see this specific issue on the GMAT, but if you perform at a high level in the Quant section, then the GMAT is more likely to throw you "quirkier" versions of rules (such as the square in this question). Ask yourself what you "know" (in this question, we know that we're dealing with a square, but we're given just 2 of the vertices) and what you "don't know" (the other 2 vertices) and work from there.

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by GMATGuruNY » Sun Feb 16, 2014 4:54 am
gmattesttaker2 wrote: Points A and C have (x,y) coordinates (-1,3) and (2,-1) respectively, and are
opposite vertices of square ABCD, what is the perimeter of square ABCD?
When no figure is given, there is a likely reason:
If a figure were given, the problem would be too EASY.
Implication:
Be careful when drawing.
The problem probably contains a TWIST.

Image
As the figure above shows, AC is the hypotenuse of a 3-4-5 triangle.
Thus, AC = 5.

AC is also the diagonal of square ABCD.
The diagonal of a square = s√2.
Since AC = 5, we get:
s√2 = 5.
s = 5/√2.

Thus:
p = 4s = 4 * (5/√2) = (4*5*√2) / (√2*√2) = (20√2)/2 = [spoiler]10√2[/spoiler].
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by gmattesttaker2 » Sun Feb 16, 2014 5:41 pm
Patrick_GMATFix wrote:Sri,

You assumed that the sides of the squares are parallel to the x and y axes. As you drawing shows, this is not possible since the shape you drew is not a square. Think of the question in more simple terms:

Given the length of a diagonal (the distance between two opposite vertices), what is the perimeter of the square? If you find the length of AC, you will have the first piece!
Image

-Patrick
Hello Patrick,

Thanks a lot for the excellent explanation (and visuals).

Best Regards,
Sri
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by gmattesttaker2 » Sun Feb 16, 2014 5:46 pm
GMATGuruNY wrote:
gmattesttaker2 wrote: Points A and C have (x,y) coordinates (-1,3) and (2,-1) respectively, and are
opposite vertices of square ABCD, what is the perimeter of square ABCD?
When no figure is given, there is a likely reason:
If a figure were given, the problem would be too EASY.
Implication:
Be careful when drawing.
The problem probably contains a TWIST.

Image
As the figure above shows, AC is the hypotenuse of a 3-4-5 triangle.
Thus, AC = 5.

AC is also the diagonal of square ABCD.
The diagonal of a square = s√2.
Since AC = 5, we get:
s√2 = 5.
s = 5/√2.

Thus:
p = 4s = 4 * (5/√2) = (4*5*√2) / (√2*√2) = (20√2)/2 = [spoiler]10√2[/spoiler].
Hello Mitch,

Thanks a lot for the excellent visual and explanation.

Best Regards,
Sri
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by gmatbestguru » Mon Feb 17, 2014 11:55 am
The two points are = A(-1,3) and C(2,-1)

Using distance formula the length of diagonal AC is =
=> sqrt(square(-1-2)+square(3+1))= sqrt(square(-3)+square(4)) = sqrt(9+16) = sqrt(25) = 5

Now, if we draw the square ABCD, ABC forms a right angled triangle.

Therefore, as per Pythagoras theorem,
=> square AC = square AB + square BC
OR square 5 = square AB + square BC
OR 25 = square AB + square BC
As ABCD is a square therefore AB = BC,therefore,
=> 25 = square AB + square AB
OR 25 = 2 * square AB
OR 25/2 = square AB
OR sqrt(25/2)= AB
OR 5/(sqrt 2)= AB

Now as denominator has a number in square root, therefore, we will rationalize the number by multiplying the numerator and denominator by "sqrt (2)"

Hence we get,

=> (5 sqrt 2)/(sqrt 2 sqrt 2) = AB = (5 sqrt 2)/2

As one side is AB, and all sides of a square are always equal, therefore, perimeter is "4AB"

=> 4 AB = 4 (5 sqrt 2)/2 = "10 sqrt 2" units Ans.

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