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Re: In the figure above, is the area of triangular region ABC
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13 Apr 2015, 01:41
Is a valid solution to this to plug in numbers instead of X to avoid painful calculations?
I simply assigned "AC" to 6 > (AC)^2 = 2(AD)^2 > 36 = 2(AD)^2. > 18 =(AD)^2. AD = \(3\sqrt{2}\)
Jumping straight to (1)+(2): So, they are Isoceles: 454590 ratio aka \(666\sqrt{2}\). Which means Area of ABC = 36/2 = 18
Area of DBA = \(6\sqrt{2} * 3\sqrt{2}\) / 2 = 18
I mean... as long as Im conistent with keeping "AC" to 6, it shouldn't really matter since i'm just expressing AC in a different way, right? I just find it way easier to work with actual numbers than with letters



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Re: In the figure above, is the area of triangular region ABC
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13 Apr 2015, 17:45
Hi erikvm, Yes, TESTing VALUES on this question is a great way to deal with a situation that is really 'technical.' As long as you're labeling your work and being thorough, you'll find that this approach works on LOTS of Quant questions. GMAT assassins aren't born, they're made, Rich
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Re: In the figure above, is the area of triangular region ABC
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14 Apr 2015, 00:31
EMPOWERgmatRichC wrote: Hi erikvm,
Yes, TESTing VALUES on this question is a great way to deal with a situation that is really 'technical.' As long as you're labeling your work and being thorough, you'll find that this approach works on LOTS of Quant questions.
GMAT assassins aren't born, they're made, Rich Thanks Rich. I am a big fan of testing values, but got a bit sceptical when it comes to DS questions.



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Re: In the figure above, is the area of triangular region ABC
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14 Apr 2015, 21:35
Hi erikvm, TESTing VALUES is a useful approach on most DS questions; you just have to be sure to adjust the tactic a bit. On a PS question, you're (typically) looking for the one answer that matches your values. On a DS question, you're TESTing multiple VALUES to see if a pattern emerges (or if you can prove that a pattern does NOT exist). Since that strategy works in so many different ways on Test Day, it's a great approach to practice throughout your studies. GMAT assassins aren't born, they're made, Rich
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Re: In the figure above, is the area of triangular region ABC
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10 May 2015, 23:07
jfk wrote: Attachment: OG13DS79v2.png In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ? (1) (AC)^2=2(AD)^2 (2) ∆ABC is isosceles. Source: OG13 DS79 Ans C. Keep CB as x. From equation 1, we get (AC)^2=2(AD)^2 , which gives as AD = x/ root 2.  NS From 2, we get both AC and CB are x. which implies AB is root 2 * x.  NS as no info about other triangle. Combine. we can find the area of both the triangles and its equal. Keeping CB as x is the key here. Hope it helps



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Re: In the figure above, is the area of triangular region ABC
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Updated on: 20 Nov 2015, 14:49
Spent too much time on this one..... Let write down what is given: Area1=Area2 ? (AC*BC)/2=(AD*AB)/2 ? >(it's easier to eliminate 2 from denominator + \(ac^2*bc^2=ad^2*ab^2\) We have also Info about Hypotenuse: \(DB^2=AD^2+AB^2\), \(AB^2=AC^2+CB^2\) 1) \(AC^2=2*AD^2\) > \(2*AD^2*bc^2\)=\(ad^2*ab^2\) > \(2*bc^2=ab^2=ac^2+bc^2 > bc^2=ac^2 ?\) Not Sufficient 2) ac=bc > ac^4=ad^2*ab^2 ?... Not Sufficient (1+2) in (1) bc^2=ac^2 ? Statement 2 answers this question directly  Yes Answer (C) Hi Math Experts, I see many people here dealing with complex numbers and roots in this problem, cano not we just square the expression in the question as I did ?
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Re: In the figure above, is the area of triangular region ABC
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20 Jul 2015, 20:19
ltkenny wrote: I still don't get it. To compare the Area of (ABC) and Area of (DBA): is 1/2(AC)(CB) = 1/2(AD)(AB) ? In the hint 1: (AC)^2=2(AD)^2 > AC = sq(2)*(AD) > AD > AC (1) Then look at the graph: (AB) is the hypotenuse of triangle (ABC) > AB > CB (2) (1) and (2) : > (AC)(CB) < (AB)(AD) > the answer is NO, their areas are not equal Thus, the answer must be A. Can anyone please explain my mistake in here? Thanks I got the same answer first. Then, i thought, AB^2=AC^2+BC^2 That means Area of triangle ABD is =.5*AB*AD =0.5*sqrt(AC^2+BC^2)*AC/sqrt(2) which maybe greater than 0.5*AC*BC . Unless you know BC and Ac values, you cannot say if area is greater, because of the square root 2, you cannot say, it reduces the value to 0.707 of the value. So, it may be or may not be greater. Hence, it's not A.



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Re: In the figure above, is the area of triangular region ABC
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28 Jul 2015, 21:04
My approach on this was:
Is Area ABC = Area DBA ?
Area ABC = (CB.AC)/2 Area DBA = (AB.AD)/2
So, we are reduced to: Is (CB.AC)/2 = (AB.AD)/2 ? even further: Is (CB.AC)= (AB.AD) ?
St1 (AC)^2=2(AD)^2 (AC)=√2(AD) (CB.√2(AD))=(AB.AD) √2(CB)= (AB) ? Cannot affirm anything about it > NOT SUFF
St2 Triangle ABC is isoceles, so (AC)=(CB), but nothing can be affirmed the other segments: (AB) and (AD) > NOT SUFF
St1/2 We know until now: √2(CB)= (AB) (AC)=(CB) (AC)^2 + (CB)^2 = (AB)^2 (Pythagorean Theorem)
(CB)^2 + (CB)^2 = (AB)^2 2(CB)^2 = (AB)^2 √2(CB)=(AB)
Going back on the first equation: (CB.AC)=(AB.AD) (CB.√2(AD))=(√2(CB).(AD)) yes, the areas are equal
I think I could've stopped in that moment where I noticed with all the three equations in hand I could find the relation between the areas.
Does this make sense ?



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Re: In the figure above, is the area of triangular region ABC
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16 Feb 2016, 19:17
took me some time to solve it.. we know that both triangles are right angles. we are asked for areas if are equal area is bxh/2
AC*CB/2 = AD*AB/2 or AC*CB=AD*AB  this is our question.
1. we know that AC^2=2AD^2. we know that: AC^2+CB^2=AB^2 2AD^2+CB^2=AB^2.
doesn't tell much.
2. ABC is isosceles. AC=CB ok, so AB^2=2AC^2 or AB=AC*sqrt(2)
area of ABC is AC^2/2
area of ABD is AB*AD/2. we know AB, but it doesn't help much, as we still have 2 unknowns  AC and AD.
1+2.
since we found out in 2 that we have 2 unknowns, 1 fills out the gap: Area of ABC=AC^2/2 or 2AD^2/2 = AD^2.
We know that AB^2=2AC^2. we also know that AC^2=2AD^2. thus, AB^2=4AD^2. or AB=2AD
Area of ABD=2AD*AD=2AD^2/2 = AD^2
we see that the areas are equal.



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Re: In the figure above, is the area of triangular region ABC
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08 Mar 2016, 06:49
I think answer should be A, official answer is C A ) IF AD= Root2*AC triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2 SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC
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Re: In the figure above, is the area of triangular region ABC
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08 Mar 2016, 06:57
vijaisingh2001 wrote: I think answer should be A, official answer is C
A ) IF AD= Root2*AC
triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2
SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC Follow posting guidelines (link in my signatures). Search for a question before you post, type in the complete question and do not use pictures for questions. Topics merged.
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Re: In the figure above, is the area of triangular region ABC
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Re: In the figure above, is the area of triangular region ABC
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08 Mar 2016, 07:07
vijaisingh2001 wrote: I think answer should be A, official answer is C
A ) IF AD= Root2*AC
triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2
SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC C is the correct answer for this question. First off, it is not AD= Root2*AC but (AC)^2= 2(AD)^2 Your statement above is NOT correct. Area (ABC) = 0.5*AC*AB = 0.5*\(\sqrt{2}\)*AD*BC = 0.7 * AD*BC Area (ADB) = 0.5*AD*AB Now although, BC < AB , do you know for sure that 0.7*BC = (a fixed value)* AB , do you know the 'fixed value'? NO. This the reason why this statement is NOT sufficient. Be careful with your interpretations in DS questions. You must be absolutely sure of the final value. Hope this helps.



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Re: In the figure above, is the area of triangular region ABC
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22 Jun 2016, 21:09
Area (ABC) = 1/2*AC*BC Area (DBA) = 1/2*DA*AB
Statement 1: AC = √2* AD. But we do not know anything about the base ISUFFICIENT
Statement 2: We just have the information about ABC INSUFFICIENT
Combining both statements: On combining, AC = CB, and AC = √2*AD AB = AC^2 + BC^2 = 2AD + 2AD = 4AD
We have the values of base and perpendicular of both triangles in terms of AD. Hence we can calculate and compare the area SUFFICIENT
Correct Option: C



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Re: In the figure above, is the area of triangular region ABC
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01 Dec 2016, 02:50
My approach on this one was as follows: I reworded the question, so if the question asks if the area of triangular region ABC is equal to the area of triangular region DBA, then it means that \(( (AC) (CB) ) / 2 = ((AD) (AB)) /2\), we can simplify it as \((AC) (CB) = (AD) (AB)\). The reworded question is then \((AC) (CB) = (AD) (AB)\) ? 1) \((AC)^2 = 2(AD)^2\) \(\sqrt[2]{(AC)^2}= \sqrt[2]{2(AD)^2}\) \(AC = AD \sqrt[2]{2}\) But we don't know anything about CB and AB so the stem is Insufficient. 2) ABC is isosceles. We don't know anything about the measures of each triangle so, stem (2) is insufficient. Together (1) and (2): We know that the proportions between AC and AD from stem 1. That's to say, \(AC = AD \sqrt[2]{2}\) And we know that an isosceles triangle has two of the sides that are equal and they always follow this proportion: \(x:x:x\sqrt[2]{2}.\) So AC and CB should be equal. That means, \(CB = AD \sqrt[2]{2}\) and if \(x = AD \sqrt[2]{2}\) then \(AB = AD \sqrt[2]{2} \sqrt[2]{2} = AD \sqrt[2]{4} = 2(AD)\) Then area of ACB is \(((AD \sqrt[2]{2}) (AD \sqrt[2]{2}) ) / 2 = (2(AD)^2)/2\) And area of DAB is \(((AD) 2(AD))/2 = (2(AD)^2)/2\) That means, \((2(AD)^2)/2 = (2(AD)^2)/2\) so area of ACB and area of DAB are equal.
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Re: In the figure above, is the area of triangular region ABC
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29 May 2017, 18:02
I think I found an effective way of solving this problem in under two minutes. Let me explain the logic of it:
For simplicity purposes, let's call \((\overline{AC})=x; (\overline{CB})=y; (\overline{AD})=w\); and \((\overline{AB})=z\).
The question is basically asking if \(xy=wz\).
Let's analyze the statements:
Statement 1)
\(x^{2}=2w^{2}\). You can solve for \(w\) and see that \(w=\frac{x\sqrt{2}}{2}\). This does not help us to answer the question, so Statement 1 is not sufficient.
Statement 2)
∆ABC being an isosceles right triangle means that the triangle is a \(x:x:x\sqrt{2}\) triangle. In other words, that \(x=y\), and consequently that \(\text{A}^{ABC}=\frac{x^{2}}{2} \rightarrow 2\text{A}^{ABC}=x^{2}\). Again, useful, but not sufficient.
Statements 1 and 2)
From Statement 2 we know that \(2\text{A}^{ABC}=x^{2}\), and from Statement 1 we know that \(x^{2}=2w^{2}\). From these, we can say that:
\(2\text{A}^{ABC}=2w^{2} \rightarrow \text{A}^{ABC}=w^{2} \rightarrow \frac{xy}{2}=w^{2}\). It's time to use some of the information we had from Statement 1, in this case regarding \(w\):
\(\frac{xy}{2}=\left(\frac{x\sqrt{2}}{2}\right)^{2} \rightarrow \frac{xy}{2}=\frac{x^{2}}{2}\).
Now, since from Statement 2 we know that \(x=y\), then the equation will hold true, and therefore, the correct answer is C.
Hope it helps.



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Re: In the figure above, is the area of triangular region ABC
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06 Jul 2017, 09:32
Area of triangle ABC= 1/2 * AC*CB Area of triangle BAD= 1/2 *AD*AB To find: AC*CB= AD*AB ? 1) AC^2= 2*AD^2 This just gives a relation between AC and AD. We can't estimate exact values. 2) AC=CB Therefore it is a 454590 triangle with sides in ratio 1:1: sq.root 2 Now our question becomes: Is AC^2= AD*AB Or Is x^2= AD*AB ? No information about any other side. Both We can say that: CB^2= 2*AD^2 AD= X/sq.root 2 Area of ABC= x^2/2 Area of DAB= x^2/2 The area of both the triangles is same.
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Re: In the figure above, is the area of triangular region ABC
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29 Apr 2018, 01:24
Hi the question is is the area of triangular region ABC equal to the area of triangular region DBA ?
from Statement 1 we know (1) (AC)^2=2(AD)^2 Which means AC is probably bigger than AD . however for the area of a triangle we need the relation to both Height and Base to draw a conclusion hence insufficient. From Statement 2 we know
(2) ∆ABC is isosceles. However we still have no idea about AD HENCE WE STILL CONT COMPARE AREAS OF BOTH THE TRIANGLES. Now together we know (1) (AC)^2=2(AD)^2 and (2) ∆ABC is isosceles.
Using both, ratio of sides of ABC are 1:1:2 √ 1:1:2 = AC:BC : AB Area of ABC = 1/2*1*1 = 1/2
Area of DAB = 1/2∗AD∗AB=1/2∗1/2 √ ∗2 √ =1/2 1/2∗AD∗AB=1/2∗1/2∗2=1/2
Areas of both the triangles is the same



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Re: In the figure above, is the area of triangular region ABC
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17 Jul 2018, 23:59
teal wrote: does anyone have any other alternative method to solve? We have to check if \(1/2 * AB * BC = 1/2* AD * AB\) i.e, \(AB * BC = AD * AB\) From (1), \(AC^2 = 2*AD^2\) \(AC = \sqrt{2}AD\) . Not Sufficient From (2), \(\triangle ABC\) is isosceles. \(AC = BC\) in \(\triangle ABC\), \(AB^2 = AC^2 + BC^2\) \(AB^2 = 2*BC^2\) . \(BC = (1/\sqrt{2})AB\). Not Sufficient From (1) and (2), \(AC * BC = \sqrt{2}AD * (1/\sqrt{2})AB\) \(AC * BC = AD * AB\) Sufficient. so (C)



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Re: In the figure above, is the area of triangular region ABC
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03 Aug 2018, 01:57
Logical Approach:
To find the Area of any right angle triangle, we need to know the value of at least two sides of that triangle.
St 1: NS Because we only know one side of each triangle. St 2: NS Because we don't have any information about another triangle.
By combining We now have the value of two sides of triangle ABC and DBA. It doesn't matter what these values are. Because we just need to know that whether we can find the area of these two triangles. In this DS question, we don't have to prove anything. Either they will be equal or not equal.
Sufficient.
Ans C




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