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505-555 Level|   Geometry|                        
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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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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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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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GMAT198917
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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@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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@Bunnel,
Can't we say area of smaller square=2*area of larger square

How?
If it is mentioned in the question then only you can consider it, by default you are not allowed to assume anything.
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@I have a quick question. In this problem, it's said that each of the vertices of the smaller square cut one side of the larger square into the ratio 1:2. In that case, it might even be that any right triangle's equation be, 'x^2 + x^2 = 10' or '2x^2 + 2x^2 = 10' since it is not given which part of the line segment of the larger square gets x and which gets 2x. In that case, my answer changes again. Then, how can I take C for an answer? Since it will not be a definite answer, my answer must be E? Bunuel
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@I have a quick question. In this problem, it's said that each of the vertices of the smaller square cut one side of the larger square into the ratio 1:2. In that case, it might even be that any right triangle's equation be, 'x^2 + x^2 = 10' or '2x^2 + 2x^2 = 10' since it is not given which part of the line segment of the larger square gets x and which gets 2x. In that case, my answer changes again. Then, how can I take C for an answer? Since it will not be a definite answer, my answer must be E? Bunuel

Here is a tip. Try constructing the case you are describing.
.

Assign angles and see what you get.
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AbdurRakib



OG Q 2017 New Question(Book Question: 184)


Attachment:
153y8e8.png

ANswer: Option C

Please check the video for the step-by-step solution.



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OG Q 2017 New Question(Book Question: 184)


Attachment:
153y8e8.png


(1) Hypotenuse of the right triangle is 10; two other sides can take several values; Insufficient.

(2) Two sides of the large square are: \(1x:2x;\) x can take numerous values; Insufficient.


Using Both information:
\(x^2+(2x)^2=10^2\); Gives specific value; Sufficient.

The answer is C
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OG Q 2017 New Question(Book Question: 184)


Attachment:
153y8e8.png
Solution:

We need to determine the side length of the larger (or circumscribed) square in centimeters.

Statement One Alone:

Knowing the side length of the smaller (or inscribed) square is not sufficient to determine the side length of the larger square since we don’t know exactly where the vertices of the smaller square are located on the respective sides of the larger square.

Statement Two Alone:

Knowing exactly the vertices of the smaller square are located on the respective sides of the larger square is not sufficient to determine the side length of the larger square since we don’t know the side length of the smaller square.

Statements One and Two Together:

We see that the vertices of the smaller square cut their respective sides of the larger square into two segments. If we let the shorter segment be x, then by statement two, the longer segment will be 2x (and notice that the side length of the larger square will then be 3x). By statement one, the side length of the smaller square is 10 cm; therefore, we can use the Pythagorean theorem on one of the unshaded triangles in the diagram to create the equation:

x^2 + (2x)^2 = 10^2

This allows us to solve for x and hence obtain the side length of the larger square. Both statements together are sufficient.

Answer: C
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