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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  [#permalink]

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New post 24 Aug 2013, 03:55
mau5 wrote:
leroyconje wrote:
WAIT WAIT WAIT....

in my opinion statement 1) is sufficient.

AC is less than AD, and BC must be less than AB (leg vs hypotenuse) therefore Area cannot be equal. 1) sufficient.

choice A is correct?


Refer to the posts above, AC>AD.


MY GOOD what a distraction!
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 18 Sep 2013, 17:14
EvaJager wrote:
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


The area of a right triangle can be easily expressed as half the product of the two legs.
Area of triangle \(ABC\) is \(0.5AC\cdot{BC}\) and that of the triangle \(DBA\) is \(0.5AD\cdot{AB}\).

The question in fact is "Is \(AC\cdot{BC} = AB\cdot{AD}\)?"

(1) From \(AC^2=2AD^2\), we deduce that \(AC=\sqrt{2}AD\), which, if we plug into \(AC\cdot{BC} = AB\cdot{AD}\), we get \(\sqrt{2}AD\cdot{BC}=AB\cdot{AD}\), from which \(\sqrt{2}BC=AB\). This means that triangle \(ABC\) should necessarily be isosceles, which we don't know.
Not sufficient.

(2) Obviously not sufficient, we don't know anything about \(AD\).

(1) and (2): If \(AC=BC=x\), then \(AB=x\sqrt{2}\), and \(AD=\frac{x}{\sqrt{2}}\).
Then \(AC\cdot{BC}=AB\cdot{AD}=x^2\).
Sufficient.

Answer C.


I'm a little confused with this method.

1) In statement 1, how can you deduce that the ABC needs to be an isosceles?
2) When you combine the statements, How do you know that \(AD=\frac{x}{\sqrt{2}}\)?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 18 Nov 2013, 21:37
teal wrote:
does anyone have any other alternative method to solve?



Don't use numbers, use logic.

Vandygrad above helped me break this down. Look up what an Isosceles triangle is: Which is a triangle that has 2 equal sides, and 2 equal angles, so with that information on hand, you know that it's a 45/45/90 , and look up how to find the area of a triangle A=Base*Height/2, or A=1/2(B*H).

The question stem states that the area of ABC and DBA, is it the same? Yes/No?

1) No values are given for ABC/or or DBA. INSUF.

So you have an equation for C, but no know lengths, so insufficient to solve for the area. Knowing the formula is irrelevant, values are important. Time Saver.

2) ABC=Isocolecs: so you know the triangle has 2 equal sides, and 2 equal angels. No lengths, to be able to determine the area of triangle ABC:

Combined) C is solved for. Given: *****ABC = Isocolesces, you have two equal sides, that and knowing that you can reduce statement 1, then use the P. Theorem to solve: Stop here. Sufficient to solve.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 07 Aug 2014, 02:04
EvaJager wrote:
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) and (2): If \(AC=BC=x\), then \(AB=x\sqrt{2}\), and \(AD=\frac{x}{\sqrt{2}}\).
Then \(AC\cdot{BC}=AB\cdot{AD}=x^2\).
Sufficient.

Answer C.


Can anyone explane the last step? AC * BC = AB * AD = x^2 ?

x * x = x√2 * x/√2

x^2 = x^2*√2/√2 ---> is this right to solve the equation like this?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 12 Mar 2015, 08:58
I understand the mathematical/algebraic approach, but is it possible to answer this question using logic alone?

For example:

Statement 1 fixes the ratio of AC to AD, but point B is still free to move in space, so insufficient.
Statement 2 fixes the ratio of AC to BC, but point D is still free to move in space, so insufficient.

Together, the ratio of all the sides are fixed, so sufficient.

Does the above approach make sense?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 12 Mar 2015, 19:53
swaggerer wrote:
I understand the mathematical/algebraic approach, but is it possible to answer this question using logic alone?

For example:

Statement 1 fixes the ratio of AC to AD, but point B is still free to move in space, so insufficient.
Statement 2 fixes the ratio of AC to BC, but point D is still free to move in space, so insufficient.

Together, the ratio of all the sides are fixed, so sufficient.

Does the above approach make sense?


The two statements are fine but you need to think through the "using both" part clearly. I have discussed the logical approach here: in-the-figure-above-is-the-area-of-triangular-region-abc-134270.html#p1134780
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 13 Apr 2015, 18:45
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.

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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 14 Apr 2015, 22:35
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.

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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 11 May 2015, 00:07
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  [#permalink]

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New post 20 Jul 2015, 21:19
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  [#permalink]

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New post 28 Jul 2015, 22:04
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  [#permalink]

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New post 16 Feb 2016, 20:17
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  [#permalink]

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New post 08 Mar 2016, 07:49
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  [#permalink]

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New post 08 Mar 2016, 07:57
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.

The list of OG2013 quant questions is at the-official-guide-quantitative-question-directory-143450.html


Do not waste your time in questioning the answers of the official guide. They are always correct.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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New post 08 Mar 2016, 08:07
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  [#permalink]

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New post 22 Jun 2016, 22:09
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  [#permalink]

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New post 01 Dec 2016, 03:50
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  [#permalink]

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New post 29 May 2017, 19:02
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  [#permalink]

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New post 06 Jul 2017, 10:32
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 45-45-90 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  [#permalink]

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New post 29 Apr 2018, 02:24
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   [#permalink] 29 Apr 2018, 02:24

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