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Re: M13 Q5 [#permalink]
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07 Nov 2011, 14:34
Bunuel wrote: sonnco wrote: Tadashi wrote: What is the area of parallelogram \(ABCD\)? 1. \(AB = BC =CD = DA = 1\) 2. \(AC = BD = \sqrt{2}\) 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 welldefined 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. Answer: C. Hope it's clear. It was the explanation that worked for me Thank you



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Re: M13 Q5 [#permalink]
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31 Jan 2012, 09:03
In a llogramOpposite sides are equal in length and opposite angles are equal in measure.so,when the diagonals are equal then it can either be a rectangle or a square, we cant find the area until we know the sides, eventually we have to go back to (1) to consider the sides =1, hence C
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Re: M13 Q5 [#permalink]
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01 Feb 2012, 03:38
thought it was B. now I know it is C. assumed even if square changes to rhombus, area would still remain same. And s2 is true only for squares And I thought Quant was my strength....drats
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Re: M13 Q5 [#permalink]
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01 Feb 2012, 03:47
Bunuel wrote: 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). Just to add to what is mentioned... It is not possible to know the sides of a rt angle triangle with only Hypotenuse is coz it can make any kind of rt. angled triangle. 454590 or 306090 or 256590. (and each shall have different L and W, imagine and see ) a Hypotenuse shall have unique set of L and W for a rt. angled triangle if its angle to the x axis was mentioned
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Re: M13 Q5 [#permalink]
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01 Feb 2012, 04:46
I choose B If we know the diagnoals is Srt 2 then its clear that its a 459045 right angle triangle hence the sides are 1,1 thus we can calculate the area using this data. Can someone explain why my approach is wrong?
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02 Feb 2012, 04:06
boomtangboy wrote: I choose B
If we know the diagnoals is Srt 2 then its clear that its a 459045 right angle triangle hence the sides are 1,1 thus we can calculate the area using this data.
Can someone explain why my approach is wrong? the diagonals are same, but the S2 doesnt say that they bisect each other. Diagonals can be anything from being at 45 or 135 deg from the xaxis to being almost 0 (flat on the ground) or almost 90 (straight up)
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31 Jan 2013, 06:32
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What is the area of parallelogram \(ABCD\)?Notice that we are told that ABCD is a parallelogram. (1) \(AB = BC =CD = DA = 1\) > all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals) or \(bh\) (where \(b\) is the length of the base and \(h\) is the altitude), so we don't have enough data to calculate the area. Not sufficient. (2) \(AC = BD = \sqrt{2}\) > the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.(1)+(2) ABCD is a rectangle and a rhombus, so it's a square > area=side^2=1^2=1. Sufficient. Answer: C.
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Re: M13 Q5 [#permalink]
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21 Feb 2013, 06:58
Bunuel wrote: What is the area of parallelogram \(ABCD\)?
Notice that we are told that ABCD is a parallelogram.
(1) \(AB = BC =CD = DA = 1\) > all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals) or \(bh\) (where \(b\) is the length of the base and \(h\) is the altitude), so we don't have enough data to calculate the area. Not sufficient.
(2) \(AC = BD = \sqrt{2}\) > the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.
(1)+(2) ABCD is a rectangle and a rhombus, so it's a square > area=side^2=1^2=1. Sufficient.
Answer: C. bro bunuel, Are the areas of a square and rhombus with sides of equal lengh different?
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Re: M13 Q5 [#permalink]
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22 Feb 2013, 01:04
Sachin9 wrote: Bunuel wrote: What is the area of parallelogram \(ABCD\)?
Notice that we are told that ABCD is a parallelogram.
(1) \(AB = BC =CD = DA = 1\) > all four sides of parallelogram ABCD are equal, which implies that ABCD is a rhombus. Area of a rhombus equals to \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals) or \(bh\) (where \(b\) is the length of the base and \(h\) is the altitude), so we don't have enough data to calculate the area. Not sufficient.
(2) \(AC = BD = \sqrt{2}\) > the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle. Area of a rectangle equals to length*width, so again we don't have enough data to calculate the area. Not sufficient. Notice that you cannot find the area of a rectangle just knowing the length of its diagonal.
(1)+(2) ABCD is a rectangle and a rhombus, so it's a square > area=side^2=1^2=1. Sufficient.
Answer: C. bro bunuel, Are the areas of a square and rhombus with sides of equal lengh different? Check here: whatistheareaofparallelogramabcd1abbccd84784.html#p1185748Hope it helps.
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Re: M13 Q5 [#permalink]
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20 Apr 2014, 08:04
1) A parallelogram whose 4 sides are equal is a rhombus. But it's just like that, nothing more so can't conclude on the area > insufficient 2) For a parallelogram, we need a height with corresponding base to know the area. Only two diagonals are not enough > insufficient Combine 2 stats: we have a rhombus  two diagonals now could be used as height and base > sufficient Choose C







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