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M13-05

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M13-05 [#permalink]

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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 \(length*width\), 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, hence \(\text{area}=\text{side}^2=1^2=1\). Sufficient.


Answer: C
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Re: M13-05 [#permalink]

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New post 25 Mar 2015, 16:53
stmnt 2: could you help me please why the (d1*d2)/2 formula cannot be used for calculating the are of a rectangle?

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New post 26 Mar 2015, 03:02
bigzoo wrote:
stmnt 2: could you help me please why the (d1*d2)/2 formula cannot be used for calculating the are of a rectangle?


First of all, two diagonals of a rectangle are equal in length so if this formula could be applied to rectangles, then it would be area=d^2/2. But this is NOT correct. Consider this: changing angles between diagonals would change its area, so a rectangle with the diagonal of d can have infinitely many different areas, depending on the angle between diagonals.

Hope it's clear.

P.S. With d^2/2 you can find the area of only one specific rectangle, namely a square: square area = d^2/2.
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Re: M13-05 [#permalink]

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New post 27 Mar 2015, 06:54
Bunuel wrote:
bigzoo wrote:
stmnt 2: could you help me please why the (d1*d2)/2 formula cannot be used for calculating the are of a rectangle?


First of all, two diagonals of a rectangle are equal in length so if this formula could be applied to rectangles, then it would be area=d^2/2. But this is NOT correct. Consider this: changing angles between diagonals would change its area, so a rectangle with the diagonal of d can have infinitely many different areas, depending on the angle between diagonals.

Hope it's clear.

P.S. With d^2/2 you can find the area of only one specific rectangle, namely a square: square area = d^2/2.


Bunuel,
So far I only knew that area of a square = (side)^2 but as per the information you have given above can we find the area if we are given the length of diagonal without really finding out the length of side in a square?

What I mean is suppose we are given length of a diagonal in a square as root 2, then do we need to find the length of sides (by using property of isoceles right triangle) or can we just do d^2/2 and find the area. I can see that both ways the area is same in this example however can you please confirm that this always works? Many thanks.

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gmatkiller88 wrote:
Bunuel wrote:
bigzoo wrote:
stmnt 2: could you help me please why the (d1*d2)/2 formula cannot be used for calculating the are of a rectangle?


First of all, two diagonals of a rectangle are equal in length so if this formula could be applied to rectangles, then it would be area=d^2/2. But this is NOT correct. Consider this: changing angles between diagonals would change its area, so a rectangle with the diagonal of d can have infinitely many different areas, depending on the angle between diagonals.

Hope it's clear.

P.S. With d^2/2 you can find the area of only one specific rectangle, namely a square: square area = d^2/2.


Bunuel,
So far I only knew that area of a square = (side)^2 but as per the information you have given above can we find the area if we are given the length of diagonal without really finding out the length of side in a square?

What I mean is suppose we are given length of a diagonal in a square as root 2, then do we need to find the length of sides (by using property of isoceles right triangle) or can we just do d^2/2 and find the area. I can see that both ways the area is same in this example however can you please confirm that this always works? Many thanks.


Yes, you can ALWAYS find the area of a square by diagonal^2/2.
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New post 27 Mar 2015, 07:01
gmatkiller88 wrote:
Bunuel wrote:
bigzoo wrote:
stmnt 2: could you help me please why the (d1*d2)/2 formula cannot be used for calculating the are of a rectangle?


Bunuel,
So far I only knew that area of a square = (side)^2 but as per the information you have given above can we find the area if we are given the length of diagonal without really finding out the length of side in a square?

What I mean is suppose we are given length of a diagonal in a square as root 2, then do we need to find the length of sides (by using property of isoceles right triangle) or can we just do d^2/2 and find the area. I can see that both ways the area is same in this example however can you please confirm that this always works? Many thanks.


Yes, you can ALWAYS find the area of a square by diagonal^2/2.


Ok. Thanks for confirming.

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Re: M13-05 [#permalink]

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hello, could you illustrate a rectangle which has both diagonals with the same length and is not a square?

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Re: M13-05 [#permalink]

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New post 23 Sep 2016, 05:43
Bunuel wrote:
Official Solution:


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 \(length*width\), 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, hence \(\text{area}=\text{side}^2=1^2=1\). Sufficient.

Answer: C



This might be a really stupid question; but if a parralelogram has all 4 sides equal, doesn't that make it a square?

Thanks

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toby001 wrote:
Bunuel wrote:
Official Solution:


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 \(length*width\), 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, hence \(\text{area}=\text{side}^2=1^2=1\). Sufficient.

Answer: C



This might be a really stupid question; but if a parralelogram has all 4 sides equal, doesn't that make it a square?

Thanks


Not necessarily, it might also be a rhombus.

Check for more here: math-polygons-87336.html
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Re: M13-05 [#permalink]

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New post 15 Feb 2017, 04:23
hello. I am really confused...((
I went for B.
Because, I thought that a parallelogram whose diagonals are equal must be a square... am i wrong?
Can someone please show/explain me a parallelogram that has the same length of diagonals other than square? thanks

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New post 15 Feb 2017, 06:45
FTR wrote:
hello. I am really confused...((
I went for B.
Because, I thought that a parallelogram whose diagonals are equal must be a square... am i wrong?
Can someone please show/explain me a parallelogram that has the same length of diagonals other than square? thanks


You did not read the solution carefully. The solution clearly says that ABCD must be a rectangle (rectangle is a parallelogram whose diagonals are equal). Not all rectangles are square.
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Re: M13-05 [#permalink]

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New post 07 May 2017, 04:46
Hi.
For (1) I think using hero's formula we can calculate area of quadrilateral given 4 sides . If then A is correct answer.

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New post 06 Dec 2017, 10:11
I think by the principle of "All triangle (regardless of shape) between 2 parallel lines and a fixed base will have equal area. Here Option A should be sufficient, the opposite sides should be parallel and the space between two parallel lines should have a equal area, when all the sides are equal.

Bunuel, Please come to rescue

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New post 06 Dec 2017, 10:15
Garv8 wrote:
I think by the principle of "All triangle (regardless of shape) between 2 parallel lines and a fixed base will have equal area. Here Option A should be sufficient, the opposite sides should be parallel and the space between two parallel lines should have a equal area, when all the sides are equal.

Bunuel, Please come to rescue


I have nothing else to add to what is already written there. From (1) we get that ABCD is a rhombus. Different rhombi have different areas.
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Re: M13-05   [#permalink] 06 Dec 2017, 10:15
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