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In quadrilateral ABCD, sides AB and BC each have length √2,
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In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
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Originally posted by imhimanshu on 04 Sep 2012, 05:06.
Last edited by Bunuel on 30 Mar 2014, 11:14, edited 1 time in total.
Added the OA.



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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04 Sep 2012, 05:43
Statement (1): We know all 4 sides. But we don't know any angle. There is more than one way to draw this quadrilateral. > not sufficient Statement (2): Again, there's one info too few. We would need one more angle or the fourth side. There is more than one way to draw this quadrilateral > not sufficient
Both statements together are sufficient. We can draw two triangles ABC and CDA. ABC is an isosceles rectangular triangle. What's important here is that CD and DA are longer than AB and BC. If this would not be the case, we could draw two quadrilaterals, one with angle CDA being smaler than 180 degrees, the other one with angle CDA being greater than 180 degrees.
We could mirror/flip triangle CDA anyways, but this way, the angle ABC would not longer be 90 degrees, but 270 degrees (the 90 degrees would then be "outside".



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Challenge Problem from Manhattan GMAT
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04 Sep 2012, 14:28
In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
I was able to solve the datasufficiency question using the 1 and 2 but the OA is E.My Approach is given below. GMAT club members kindly check and let me know whether there is anything wrong with my approach.
Area of quadrilateral=1\2 *d1(p1+p2) d1=lenght of one digaonal;p1=lenght of perpendicular from one vertex.p2=lenght of perpendicular from opp vertex to the diagonal. 1.Using 1. In traingle ADC of quadrilateral ABCD AD=DC. Draw perpendicular say DE from vertex D onto diagonal AC of quadrilateral.Now traingles ADE and CDE are congruent traingles AAS congruency(<DAE=<DCA ;<DEA=<DEC=90 ;AD=DC), so AE=EC.But we cant calculate length of diagonal AC from the info, hence it is insufficient.
2.Using this info we can calculate the length of diagonal AC from pythagoras theorem.Lenghth of other perpendicular drawn from vertex B on AC say BF can be found out using the same procedure as above ie by AAS congruency.But we cant calculate lenght of other perpendicular drawn from vertex D with this infro.Hence this is aslo insufficient.
Using (1)+(2) we get Length of diagonal and lenght of perpendicular DE and BF (E and F will be the same point as from AAS congruency of two traingles we get AE+EC =AC and AF=FC=AC).Hence both together are sufficeint to answer the problem.
Let me know your views on this problem.



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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05 Sep 2012, 14:47
I believe it's E.
Statement 1 is not sufficient because we don't know if angle ABD is acute or obtuse.
Statement 2 is not sufficient because of aforementioned reasons.
Even with both statements we still have nothing about the angle ABD.



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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07 Sep 2012, 04:44
if we know four sides of the quadrilateral we can find its area...
s=a+b+c+d/4
area =sqrt of s*(sa)*(sb)*(sc)*(sd) ....
so answer A ...
Am i missing something.....



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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10 Sep 2012, 11:57
i guess with both, statements 1 and 2 together, you are able to determine the area of the given quadrilateral. please refer to the image attached.. Attachment:
quadABCD.jpg [ 19.47 KiB  Viewed 13922 times ]
you can see that once you have the 2nd condition, you can find the diagonal AC of the quad. and when you calculate the length of the diagonal, you end up getting an equilateral triangle ADC on the other side. So, 1 you have a right triangle and 2 you have an equilateral triangle, the areas of both of which can be calculated. Thus, I would go with option C.
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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11 Sep 2012, 12:42
imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. I go with answer 'D' as both the sentences are sufficient to give the answer.



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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10 Oct 2012, 05:21
imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. From stem AB = BC = 2^1/2 CD = 2 1) AD = 2 > This can form a KITE if we knew the angles but since there is no info about the angles included >Insufficient 2) One of the triangle is right angle triangle but no info about the shape of other triangle >Insufficient. 1+2) If we combine both we get KITE ( a combination of 2 Isosceles Right angle triangles) > Sufficient Answer C Note : Not able to include the diagram, you have to imagine.
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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10 Oct 2012, 10:59
imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. To the best of my knowledge, on the GMAT only convex polygons are considered, except maybe the cases when a drawing is supplied. This is just another example when Manhattan is going overboard.
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. I'm going with C here. Please provide OA ok? Here's my reasoning Statement 1 we have that AD which is the diagonal of the right triangle formed in the polygon is 2 therefore knowing two sides and one angle is sufficient to know the measures of the right triangle but thing is we don't know about BC is this side perpendicular to AB or not? Insufficient Statement 2 only tells us that the angle in ABC is 90 degrees but still no way to find the area of the other right triangle. Now with both statements together we know the area of the right triangle and we know that AB is perpendicular to BC hence the area of the given square is 2 which added to the area of the right triangle can provide the total area of the figure Answer is thus C Please let me know if this is sound reasoning Cheers J
Originally posted by jlgdr on 30 Mar 2014, 09:17.
Last edited by jlgdr on 07 Apr 2014, 05:51, edited 1 time in total.



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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30 Mar 2014, 11:16
jlgdr wrote: imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. I'm going with A here. Please provide OA ok? Here's my reasoning Statement 1 we have that AD which is the diagonal of the right triangle formed in the polygon is 2 therefore knowing two sides and one angle is sufficient to know the measures of the right triangle but thing is we don't know about BC is this side perpendicular to AB or not? Insufficient Statement 2 only tells us that the angle in ABC is 90 degrees but still no way to find the area of the other right triangle. Now with both statements together we know the area of the right triangle and we know that AB is perpendicular to BC hence the area of the given square is 2 which added to the area of the right triangle can provide the total area of the figure Answer is thus C Please let me know if this is sound reasoning Cheers J The correct answer is E: Attachment:
quadrophenia3.jpg [ 9.69 KiB  Viewed 12797 times ]
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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17 Apr 2014, 11:36
Bunuel wrote: jlgdr wrote: imhimanshu wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
Please post your reasoning. I'm going with A here. Please provide OA ok? Here's my reasoning Statement 1 we have that AD which is the diagonal of the right triangle formed in the polygon is 2 therefore knowing two sides and one angle is sufficient to know the measures of the right triangle but thing is we don't know about BC is this side perpendicular to AB or not? Insufficient Statement 2 only tells us that the angle in ABC is 90 degrees but still no way to find the area of the other right triangle. Now with both statements together we know the area of the right triangle and we know that AB is perpendicular to BC hence the area of the given square is 2 which added to the area of the right triangle can provide the total area of the figure Answer is thus C Please let me know if this is sound reasoning Cheers J The correct answer is E: Attachment: quadrophenia3.jpg Hi Bunnel, When I use st1 and st2 I get one 454590 triangle. and another one an equilateral triangle with site 2. I can get area of quadrilateral by adding both the areas. I choose C but official ans is E. Please clarify. Thanks



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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18 Apr 2014, 02:45
PathFinder007 wrote: Bunuel wrote: jlgdr wrote: In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°.
I'm going with A here. Please provide OA ok?
Here's my reasoning
Statement 1 we have that AD which is the diagonal of the right triangle formed in the polygon is 2 therefore knowing two sides and one angle is sufficient to know the measures of the right triangle but thing is we don't know about BC is this side perpendicular to AB or not? Insufficient
Statement 2 only tells us that the angle in ABC is 90 degrees but still no way to find the area of the other right triangle.
Now with both statements together we know the area of the right triangle and we know that AB is perpendicular to BC hence the area of the given square is 2 which added to the area of the right triangle can provide the total area of the figure
Answer is thus C
Please let me know if this is sound reasoning Cheers J The correct answer is E: Attachment: quadrophenia3.jpg Hi Bunnel, When I use st1 and st2 I get one 454590 triangle. and another one an equilateral triangle with site 2. I can get area of quadrilateral by adding both the areas. I choose C but official ans is E. Please clarify. Thanks Not sure how to clarify better than the diagram: As you can see both those figure satisfy all the conditions and have different areas. Thus the answer must be E. Hope it helps.
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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19 Apr 2014, 01:30
Thanks for clarification. But i think second figure is not a convex polygon(because interior ABC angle is 270 and not less than 180). And in GMAT I guess we are concerned about convex polygon only? And if we agree that second figure is valid, then can we say ans is C? Also i feel is it really possible to draw such diagram with precision in exam ? Any alternative way to build the thought process for exam? Thanks. Manoj Parashar



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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21 Apr 2014, 05:09
manojISB wrote: Thanks for clarification. But i think second figure is not a convex polygon(because interior ABC angle is 270 and not less than 180). And in GMAT I guess we are concerned about convex polygon only? And if we agree that second figure is valid, then can we say ans is C? Also i feel is it really possible to draw such diagram with precision in exam ? Any alternative way to build the thought process for exam? Thanks. Manoj Parashar Yes, OG defines a quadrilateral as a polygon with four sides. Next, it says that the term "polygon" will be used to mean a convex polygon, that is, a polygon in which each interior angle has a measure of less than 180°. Hence this question violates GMAT definition of a quadrilateral, which makes the question/solution presented by MGMAT flawed. I wouldn't worry about this question at all.
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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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29 Jun 2014, 11:09
Is there a set of details that we must know in order to find the area of a quadrilateral?



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In quadrilateral ABCD, sides AB and BC each have length √2,
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24 Sep 2015, 03:35
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD? (1) The length of side AD is 2. (2) The angle between side AB and side BC is 90°. In the original condition, there are 5 variables (4 sides and 1 diagonal line) and 3 equations(sides AB and BC each have length √2, while side CD has length 2) and thus we need 2 more equations to match the number of variables and equations. Since there is 1 equation each in 1) and 2), the best answer is C by our DS definition. However, this type of question always shows the diagram in actual exam. If diagram is not shown, E is the answer because we don't know the alphabetical orders of vertices (ABCD? ACBD? BDAC?) Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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In quadrilateral ABCD, sides AB and BC each have length √2,
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30 Mar 2018, 08:28
Zinsch123 wrote: Statement (1): We know all 4 sides. But we don't know any angle. There is more than one way to draw this quadrilateral. > not sufficient Statement (2): Again, there's one info too few. We would need one more angle or the fourth side. There is more than one way to draw this quadrilateral > not sufficient
Both statements together are sufficient. We can draw two triangles ABC and CDA. ABC is an isosceles rectangular triangle. What's important here is that CD and DA are longer than AB and BC. If this would not be the case, we could draw two quadrilaterals, one with angle CDA being smaler than 180 degrees, the other one with angle CDA being greater than 180 degrees.
We could mirror/flip triangle CDA anyways, but this way, the angle ABC would not longer be 90 degrees, but 270 degrees (the 90 degrees would then be "outside". Though it is a old post. Can anyone plz help with the logic of the above explanation (especially the highlighted part). Thanks



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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13 Apr 2018, 22:44
Zinsch123 wrote: Statement (1): We know all 4 sides. But we don't know any angle. There is more than one way to draw this quadrilateral. > not sufficient Statement (2): Again, there's one info too few. We would need one more angle or the fourth side. There is more than one way to draw this quadrilateral > not sufficient
Both statements together are sufficient. We can draw two triangles ABC and CDA. ABC is an isosceles rectangular triangle. What's important here is that CD and DA are longer than AB and BC. If this would not be the case, we could draw two quadrilaterals, one with angle CDA being smaler than 180 degrees, the other one with angle CDA being greater than 180 degrees.
We could mirror/flip triangle CDA anyways, but this way, the angle ABC would not longer be 90 degrees, but 270 degrees (the 90 degrees would then be "outside". please explain reason for option C to be incorrect, if possible with pictures



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Re: In quadrilateral ABCD, sides AB and BC each have length √2,
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14 Apr 2018, 00:46
Ritz1605 wrote: Zinsch123 wrote: Statement (1): We know all 4 sides. But we don't know any angle. There is more than one way to draw this quadrilateral. > not sufficient Statement (2): Again, there's one info too few. We would need one more angle or the fourth side. There is more than one way to draw this quadrilateral > not sufficient
Both statements together are sufficient. We can draw two triangles ABC and CDA. ABC is an isosceles rectangular triangle. What's important here is that CD and DA are longer than AB and BC. If this would not be the case, we could draw two quadrilaterals, one with angle CDA being smaler than 180 degrees, the other one with angle CDA being greater than 180 degrees.
We could mirror/flip triangle CDA anyways, but this way, the angle ABC would not longer be 90 degrees, but 270 degrees (the 90 degrees would then be "outside". please explain reason for option C to be incorrect, if possible with pictures As Bunuel has already highlighted the problem with option C by using below diagram I will be further explaining it to you. Attachment:
Ritz.PNG [ 36.98 KiB  Viewed 774 times ]
First we should connect AC. AC= 2 , Using pythagoras theorem in \(\triangle\)ABC As per Figure 1 : Area of Quad ABCD = Area of \(\triangle\)ABC + Area of \(\triangle\)ADC As per Figure 2 : Area of Quad ABCD = Area of \(\triangle\)ADC  Area of \(\triangle\)ABC So Combining Statement 1 and Statement 2 is not giving a single value of Area of Quad ABC , so OA should be E
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