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What is the area of parallelogram \(ABCD\)? (1) \(AB = BC = CD = DA = 1\) (2) \(AC = BD = \sqrt{2}\)
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16 Sep 2014, 00:48
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
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Re: M1305
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25 Mar 2015, 17:53
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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26 Mar 2015, 04:02
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: M1305
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27 Mar 2015, 07:54
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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27 Mar 2015, 07:57
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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27 Mar 2015, 08:01
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: M1305
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05 Aug 2016, 17:15
hello, could you illustrate a rectangle which has both diagonals with the same length and is not a square?



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Re: M1305
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23 Sep 2016, 06:43
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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Re: M1305
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23 Sep 2016, 06:49
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: mathpolygons87336.html
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15 Feb 2017, 05:23
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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Re: M1305
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15 Feb 2017, 07:45
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: M1305
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07 May 2017, 05:46
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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07 May 2017, 07:57
anupsk wrote: Hi. For (1) I think using hero's formula we can calculate area of quadrilateral given 4 sides . If then A is correct answer. A rhombus with a side of 1 can have infinitely many areas. So, no formula will help.
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06 Dec 2017, 11:11
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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06 Dec 2017, 11:15
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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10 Mar 2018, 16:19
No formulas, calculations or even notes are required here  it can be solved visibly in about 30 seconds with basic geometry rules Remember that a square is also a rectangle, a parallelogram and a rhombus, and a rectangle is also a parallelogram. To maximise the area for any given perimeter, form a square
(1) tells us the perimeter, and that the sides are equal. We can make various rhombus shapes from this information, from a square (max area) to a very narrow diamond (smaller area). Minimum area is zero, when we squash the diamond so flat it becomes a line. In any case, there are many possible areas so we cannot determine. Insufficient (II) This information tells us the diagonals are equal. This means that the parallelogram is in fact a square or a rectangle (geometry rules). Insufficient
(i) and (ii)  we have must have square with sides equal to 1 and diagonals equal to rt2. Area is 1. Sufficient
Basically 1 tells us we that the parallelogram that is a rhombus, and 2 tells us that that the parallelogram that is a rectangle. Combining them, we must have a square, as only a square is both a rectangle and a rhombus. We actually only need the diagonal or the side value to calculate the area in this case
Remember the rules  Square  sides and diagonals equal Rectangle  diagonals equal Rhombus  sides equal All are parallelograms



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tinyinthedesert wrote: No formulas, calculations or even notes are required here  it can be solved visibly in about 30 seconds with basic geometry rules Remember that a square is also a rectangle, a parallelogram and a rhombus, and a rectangle is also a parallelogram. To maximise the area for any given perimeter, form a square
(1) tells us the perimeter, and that the sides are equal. We can make various rhombus shapes from this information, from a square (max area) to a very narrow diamond (smaller area). Minimum area is zero, when we squash the diamond so flat it becomes a line. In any case, there are many possible areas so we cannot determine. Insufficient (II) This information tells us the diagonals are equal. This means that the parallelogram is in fact a square or a rectangle (geometry rules). Insufficient
(i) and (ii)  we have must have square with sides equal to 1 and diagonals equal to rt2. Area is 1. Sufficient
Basically 1 tells us we that the parallelogram that is a rhombus, and 2 tells us that that the parallelogram that is a rectangle. Combining them, we must have a square, as only a square is both a rectangle and a rhombus. We actually only need the diagonal or the side value to calculate the area in this case
Remember the rules  Square  sides and diagonals equal Rectangle  diagonals equal Rhombus  sides equal All are parallelograms While I appreciate your insight(I actually never thought of the answer this way) statement one can not have several different quadrilaterals with different areas.If we were told only the perimeter and that it was a quadrilateral we could indeed form several quadrilaterals of several different areas. The square would have the largest. But we are not only told the area of the quadrilateral but also told how the perimeter should be distributed : that is so that each side is equal. There are only two such shapes possible : the square and the rhombus.Every other quadrilateral with perimeter 4 will not have all sides equal. This knowledge applies to triangles as well only that there is only one possible triangle with equal sides: what we call the equilateral
This still makes statement 1 insufficient










