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In the quilting pattern shown above, a small square has its vertices o

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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 01 Jul 2016, 05:43
3
28
00:00
A
B
C
D
E

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  25% (medium)

Question Stats:

76% (01:30) correct 24% (01:42) wrong based on 554 sessions

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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 01 Jul 2016, 08:53
5
1
1 is clearly insufficient.
2. the ratio is given as 1:2 again insufficient.
combining both
using Pythagoras theorem.
10^2=x^2+(2x)^2
100= 5x^2
this should give us x followed by the length of the side of the larger square.
Therefore C
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 01 Jul 2016, 08:10
C

1)side of small sq is given but no info about large sq. Insuff
2)side of small sq not given and info about large sq is there. insuff

suppose side of large sq is 3x so its divided into x and 2x and side of small square is given(which acts as hypotenuse of all4 triangles)

apply pythagos. theorem and get value of x
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 02 Jul 2016, 12:25
1
There is one unknown, that is the side of the larger square.
We need something that gives us the side of the bigger square that is in either a variable form which could be determined or in a direct value form.
Statement 1 does not have any relation to side of bigger square, hence insufficient.
Statement 2 gives a variable relation as, if side of bigger square is 3x, then the triangles whose hypotenuse is given by Statement 1 has sides 2x and x which gives an equation to determine x (4x^2 + X^2 = 10^2).
Therefore both statement are required together to arrive at a definite answer.
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New post 17 Feb 2017, 05:30
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.
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New post 21 May 2017, 14:29
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.



Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.
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New post 21 May 2017, 20:17
4
pafrompa wrote:
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.



Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.


You are saying yourself that you assumed that it must be 30-60-90 triangle. Why? Do you have any reasons supporting this claim? No.

So, you assumed with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=10\).

Hope it's clear.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 22 May 2017, 17:15
Bunuel wrote:
pafrompa wrote:
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.



Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.


You are saying yourself that you assumed that it must be 30-60-90 triangle. Why? Do you have any reasons supporting this claim? No.

So, you assumed with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=10\).

Hope it's clear.


Clear as day, thank you. I guess sometimes I force myself to try to see things that aren't there with DS. Thanks Bunuel
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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New post 08 Apr 2019, 22:34
@Bunnel,

Why can't we have answer as 'A'
My Intake:

st1: Side length of smaller square is 10 cm.
Can't we say area of smaller square=2*area of larger square
10*10=2 S^2
S^2=50
S=root 50

Is this possible?
Please help!!!
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Re: In the quilting pattern shown above, a small square has its vertices o   [#permalink] 08 Apr 2019, 22:34
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