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M13 Q5

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M13 Q5 [#permalink]  03 Sep 2008, 16:35
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What is the area of parallelogram $$ABCD$$?

1. $$AB = BC =CD = DA = 1$$
2. $$AC = BD = \sqrt{2}$$

SOLUTION IS HERE: m13-q5-69732-20.html#p1176059

[Reveal] Spoiler:
Source: GMAT Club Tests - hardest GMAT questions

S_2 is not sufficient. By reducing side AB we can make the area of the parallelogram arbitrarily small.

From S_1 + S_2 it follows that ABCD is a well-defined square. Its area equals .

The correct answer is C.

I thought that the answer is B, since if AC and BD are equal, and since the shape is a parallelogram then it would be square making the area findable?

Can anyone explain?
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Re: M13 Q5 [#permalink]  03 Sep 2008, 20:03
"if AC and BD are equal, and since the shape is a parallelogram then it would be square" is an incorrect deduction. Diagonals of equal length will convert a parallelogram to a rectangle, not a square. C makes sense
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M013 #5 [#permalink]  03 Oct 2008, 12:39
What is the area of parallelogram ABCD ?

1) AB=BC=CD=DA=1
2) AC=BD = Square root of 2

S1 is not sufficient. By reducing angle ABC we can make the area of the parallelogram arbitrarily small.

S_2 is not sufficient. By reducing side AB we can make the area of the parallelogram arbitrarily small.

From S_1 + S_2 it follows that ABCD is a well-defined square. Its area equals 1*1*1.

My question please: Why doesn't S1 alone tell us that the par. is a square and that it's area is 1? If we were to decrease the abc angle, as S1 explains, then the other angles would change and the length would no longer be 1. Please explain. Thanks.
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Re: M013 #5 [#permalink]  04 Oct 2008, 07:28
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Look at this lift:

See the square (or almost square) right in the middle? Imagine how that changes shape as the lift goes up and down. The angles change, but the sides are steel bolted together so the lengths cannot change. As the lift goes up from the point where it makes a square in the middle, the area decreases until, theoretically, all the sides line up and area is 0. Same with going down from the mid-point.
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Re: M013 #5 [#permalink]  19 Apr 2009, 21:54
If parallelograms have equal sides, why wouldn't S_2 be sufficient? If diagonals are equal, the object must be a square.

Thanks.
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Re: M013 #5 [#permalink]  20 Apr 2009, 17:03
fif4u wrote:
If parallelograms have equal sides, why wouldn't S_2 be sufficient? If diagonals are equal, the object must be a square.

Thanks.

I don't think that's true. The diagonals are equal even in a rectangle

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Re: M13 Q5 [#permalink]  22 Jan 2010, 05:20
You always refer to rectangle in order to vertices. Therefore if the parallelogram is ABCD then AC and BD are digonals. This assumption is acceptable in GMAT standard.
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Re: M13 Q5 [#permalink]  22 Jan 2010, 06:37
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when all the sides of a parallelogram and its diagonal are equal than that parallelogram is square

choice 1 doesn't have sufficient information it can either be square or rhombus in that case area will differ
choice 2 same as choice 1

but by knowing both choices we can identify the area

so my ans is C
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Re: M13 Q5 [#permalink]  22 Jan 2010, 07:36
At first glance I thought A. We cannot get the area from a parellogram by lxh since we don't actually know the height. Consider
. We know the base is length one and the lenth of a side as 1. But we don't actually know the height. For some reason I thought it was a square with all 4 sides being of equal length. But you don't actually know it is a square without first knowing the diagonals are equal in length. So you need both S1 and S2 to get the area.
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Re: M013 #5 [#permalink]  22 Jan 2010, 08:30
viperm5 wrote:
fif4u wrote:
If parallelograms have equal sides, why wouldn't S_2 be sufficient? If diagonals are equal, the object must be a square.

Thanks.

how do you even know they're diagonals?
doesnt that depend on how you draw the parallalogram? the points are arbitrary no?

You always refer to rectangle in order of vertices. Therefore if the parallelogram is ABCD then AC and BD are digonals. This assumption is acceptable in GMAT standard.
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Re: M13 Q5 [#permalink]  23 Jan 2010, 00:38
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varun2410 wrote:
chrischrischris wrote:
At first glance I thought A. We cannot get the area from a parellogram by lxh since we don't actually know the height. Consider
. We know the base is length one and the lenth of a side as 1. But we don't actually know the height. For some reason I thought it was a square with all 4 sides being of equal length. But you don't actually know it is a square without first knowing the diagonals are equal in length. So you need both S1 and S2 to get the area.

But if we know all 4 side of parallelogram we can find area without knowing hight.as shown in fig.we cut triangle portion and put on other side we allways get rectangle or square . is it true? please tell me if i am wrong ?

Yes, you're right and wrong. If we cut the part and stick it to the other side we'll get a RECTANGLE the base of which will be the same as that of paralelogram, but the hight will be smaller. The more you get it lower, the smaller the area becomes.
That is why a square has the greatest area with the same perimter comparing to other quadrilaterals.
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Re: M13 Q5 [#permalink]  23 Jan 2010, 21:09
What is the area of parallelogram $$ABCD$$?

1. $$AB = BC =CD = DA = 1$$
2. $$AC = BD = \sqrt{2}$$

I thought that the answer is B, since if AC and BD are equal, and since the shape is a parallelogram then it would be square making the area findable?

Can anyone explain?

Yes - C is correct. B is not sufficient because you need to know the length of the adjacent sides to determine the shape of the parallelogram or rather the angles.

combine 1 and 2 will ensure that it is a square thus give you the definite answer for the area. Otherwise, Area= 2* AB*BC*sine (B).
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Re: M13 Q5 [#permalink]  26 Jan 2011, 06:08
i dont thinnk anyone has any doubt about it.
everything is well explained above.
if anyone has any doubt please post here.
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Re: M13 Q5 [#permalink]  26 Jan 2011, 07:07
Rhombus & Suare both have same sides, same diagonals, both are parallelograms.

But both have different areas ... Thats why we need both statements ...
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Re: M13 Q5 [#permalink]  26 Jan 2011, 08:15
area of parallellgram = 2* sqrt [ s(s-a)(s-b)(s-d)]

where a & b are two adjacent sides and d is the diagonal

so ans to the question can be found only by combining two statements together
Hence" C" is the correct ans
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Re: M13 Q5 [#permalink]  27 Jan 2011, 14:59
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Expert's post
sonnco wrote:
What is the area of parallelogram $$ABCD$$?

1. $$AB = BC =CD = DA = 1$$
2. $$AC = BD = \sqrt{2}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

S_2 is not sufficient. By reducing side AB we can make the area of the parallelogram arbitrarily small.

From S_1 + S_2 it follows that ABCD is a well-defined square. Its area equals .

The correct answer is C.

I thought that the answer is B, since if AC and BD are equal, and since the shape is a parallelogram then it would be square making the area findable?

Can anyone explain?

I really don't understand how statement 2 is not sufficient.

If we know the parallelogram diagonals are equal the figure can be either a square or a rectangle. We are given the diagonal length of \sqrt{2}. If the diagonals are equal doesn't it make the the 4 angles inside the figure 90 degrees each?

Isosceles right triangle 1,1,\sqrt{2}

Area = 1?

If my reasoning is off please correct me or PM please. I am stumped.

Statement (2) says that the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle (square is just a special case of a rectangle). But knowing the length of the diagonals of a rectangle is not enough to calculate its area. So this statement is not sufficient.

Complete solution:

What is the area of parallelogram $$ABCD$$?

(1) $$AB = BC =CD = DA = 1$$ --> ABCD is a rhombus. Area of rhombus d1*d2/2 (where d1 and d2 are the lengths of the diagonals) or b*h (b is the length of the base, h is the altitude). Not sufficient.

(2) $$AC = BD = \sqrt{2}$$ --> ABCD is a rectangle. Area of a rectangle L*W (length*width). Not sufficient.

(1)+(2) ABCD is rectangle and rhombus --> ABCD is square --> Area=1^2=1. Sufficient.

Hope it's clear.
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Re: M13 Q5 [#permalink]  28 Jan 2011, 03:23
abhijain020786 wrote:
why cant answer be b.
b says diagonals length is equal that means it can be either square or rectangle.
now the length of diagonal is \sqrt{2}....now in both the square and rectangle...take sides AD and AB..these sides should follow pythagoras theorem with diagonal DB... so
DB^2= AB^2 + AD^2
2 = AB^2 + AD^2
which means AB and AD can be equal only then u can get 2. if i am wrong then can someone suggest some other value for the equation to get 2.

your answer would have been true, only when it was mentioned that sides are positive integer.
since no such condition was given, you can't say that its true.

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Re: M13 Q5 [#permalink]  29 Jan 2011, 09:26
im sry to nag about this question, but its not clear for me.

if i know that the Diagonal lines are cutting each other to half (in a parallelogram - they cut each other to half)

and i know each one of them = sqr 2 i can know the areas of the triangles...

can someone show me with a picture why i am wrong? i cannot c the pictures uploaded before.

thanks guys.
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Re: M13 Q5 [#permalink]  29 Jan 2011, 09:40
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144144 wrote:
im sry to nag about this question, but its not clear for me.

if i know that the Diagonal lines are cutting each other to half (in a parallelogram - they cut each other to half)

and i know each one of them = sqr 2 i can know the areas of the triangles...

can someone show me with a picture why i am wrong? i cannot c the pictures uploaded before.

thanks guys.

You can not find the area of a rectangle just knowing the length of its diagonal. Area of a rectangle is length*width and knowing the length of diagonal is not enough to calculate these values. In other words knowing the length of hypotenuse (diagonal) in a right triangle (created by length and width), is not enough to find the legs of it (length and width).
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Re: M13 Q5 [#permalink]  21 Jun 2011, 07:54
I read Bunuel explanation above.

But its still not clear to me y B cant be sufficent ?

if diagonals are root 2. Does this not imply we have a square with sides 1,1 ? since root 2 in the hypotenuse of a 45-90-45 triangle ? so it cant be a rectangle. It has to be a square- sufficent

@karishma/Fluke. does the logic in the post below apply to the problem at hand as well ?

perimeter-of-a-triangle-57595.html
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Re: M13 Q5   [#permalink] 21 Jun 2011, 07:54

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