GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 Jan 2020, 06:26 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the figure above, is the area of triangular region ABC

Author Message
TAGS:

### Hide Tags

Intern  Joined: 17 Sep 2011
Posts: 43
GMAT 1: 660 Q43 V38
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

17
192 00:00

Difficulty:   75% (hard)

Question Stats: 62% (02:25) correct 38% (02:22) wrong based on 2511 sessions

### HideShow timer Statistics In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(2) ∆ABC is isosceles.

Source: OG13 DS79

Attachment: OG13DS79v2.png [ 10.04 KiB | Viewed 80145 times ]
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10021
Location: Pune, India
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

39
23
teal wrote:
does anyone have any other alternative method to solve?

You can solve sufficiency questions of geometry by drawing some diagrams too.

Attachment: Ques3.jpg [ 4.81 KiB | Viewed 75542 times ]

We need to compare areas of ABC and DAB. Notice that given triangle ABC with a particular area, the length of AD is defined. If AD is very small, (shown by the dotted lines) the area of DAB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.

Now look at the statements:

$$AD = AC/\sqrt{2}$$
The area of ABC is decided by AC and BC, not just AC. We can vary the length of BC to see the relation between AC and AD is not enough to say whether the areas will be the same (see diagram). So insufficient.

Attachment: Ques4.jpg [ 7.15 KiB | Viewed 75647 times ]

(2) ∆ABC is isosceles.

Using both, ratio of sides of ABC are $$1:1:\sqrt{2}$$ = AC:BC : AB
Area of ABC = 1/2*1*1 = 1/2

Area of DAB = $$1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2$$

Areas of both the triangles is the same
_________________
Karishma
Veritas Prep GMAT Instructor

Director  Joined: 22 Mar 2011
Posts: 583
WE: Science (Education)
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

40
18
jfk wrote:
Attachment:
OG13DS79v2.png
In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(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.

_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
##### General Discussion
Senior Manager  Joined: 13 Jan 2012
Posts: 272
Weight: 170lbs
GMAT 1: 740 Q48 V42
GMAT 2: 760 Q50 V42
WE: Analyst (Other)
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

30
17
I started with
Area = .5bh
Area(ABC) = .5*AC*CB

So we need to know whether AC*CB = AD*AB

This tells us the relationship between AC and AD but not enough to derive the full areas. We need to know about CB and AB.

Insufficient.

BCE

2) ∆ABC is isosceles.

This tells us that AC = CB, and furthermore, we then know AB, but nothing about AD.

Insufficient.

1+2) Here, we know the relationship between AC, AD, and CB explicitly. We can also determine AB.

Sufficient. Remember, we don't need to derive the area. Just determine whether or not we can determine if they will be equal.

Intern  Joined: 17 Sep 2011
Posts: 43
GMAT 1: 660 Q43 V38
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

2

After some more calculations I found the following solution, which is similar to yours:

Area of ABC = Area of DBA?
0.5 AC*CB = 0.5 = AD*AB ?

Use Pythagorean theorem to simplify further:

$$AB^2=AC^2+CB^2$$
$$AB=\sqrt{AC^2+CB^2}$$

Therefore, the question reduces to:

$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$

(1) $$AC = \sqrt{2}*AD$$

So we can get rid of AC. Still need AD, AB and CB though:

$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$
$$\sqrt{2}*AD*AB = AD*\sqrt{AC^2+CB^2}?$$
$$\sqrt{2}*AB =\sqrt{2*AD^2+CB^2}?$$

We cannot simplify or reduce further. Insufficient.

(2) ABC = isosceles means AC = CB

So we can replace AC:

$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$
$$CB*AB = AD*\sqrt{2*CB^2}?$$
$$CB*AB = AD*\sqrt{2}*CB?$$
$$AB = AD*\sqrt{2}?$$

We cannot simplify or reduce further. Insufficient.

(1) and (2)

(1) actually answers the question we are left with in (2):

$$AB = AD*\sqrt{2}?$$

I am not sure if I could have done this in 2 mins.

Key take aways:

- Rephrase and simplify question as much as possible, given known formulas etc.
- To check C: use work you do for checking (1) also for checking (2) and vice versa.
- Don't get thrown off my complicated formulas Hope it helps others. Please let me know if I made any mistake!

Thanks a lot!
Manager  Joined: 13 May 2010
Posts: 101
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

does anyone have any other alternative method to solve?
Intern  Joined: 06 Apr 2011
Posts: 13
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

Nice and succinct explanation Eva...
Intern  Joined: 04 Sep 2012
Posts: 3
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

I still don't get it.
To compare the Area of (ABC) and Area of (DBA):
Then look at the graph:
(AB) is the hypotenuse of triangle (ABC)
-> AB > CB (2)
(1) and (2) :
-> the answer is NO, their areas are not equal
Thus, the answer must be A.
Can anyone please explain my mistake in here? Thanks
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10021
Location: Pune, India
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

1
ltkenny wrote:
I still don't get it.
To compare the Area of (ABC) and Area of (DBA):
Then look at the graph:
(AB) is the hypotenuse of triangle (ABC)
-> AB > CB (2)
(1) and (2) :
-> the answer is NO, their areas are not equal
Thus, the answer must be A.
Can anyone please explain my mistake in here? Thanks

$$AD = AC/\sqrt{2}$$
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 04 Sep 2012
Posts: 3
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

Ah, I get it. Thank you for pointing my error.
Intern  Joined: 21 Jul 2013
Posts: 3
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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?
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 576
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

1
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.
_________________
Intern  Joined: 21 Jul 2013
Posts: 3
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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!
Manager  Joined: 15 Aug 2013
Posts: 226
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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 ?

(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.

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 $$AD=\frac{x}{\sqrt{2}}$$?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10021
Location: Pune, India
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

1
russ9 wrote:
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 ?

(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.

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 $$AD=\frac{x}{\sqrt{2}}$$?

In statement 1, the question boils down to: Is $$\sqrt{2}BC=AB$$?
or Is $$BC/AB=1/\sqrt{2}$$?

In a right triangle, the ratio of a leg and hypotenuse will be 1:\sqrt{2} if the third side is also 1 i.e. only if the triangle is isosceles. You can figure this from pythagorean theorem
$$1^2 + x^2 = \sqrt{2}^2$$
$$x = 1$$

On combining the statements, we know that ABC is isosceles so AC = BC. So ratio of sides $$AC:BC:AB = 1:1:\sqrt{2}$$i.e. the sides are $$x, x$$ and $$\sqrt{2}x$$
From statement 1 we know that $$AC = \sqrt{2}AD$$
So $$AC = x = \sqrt{2}AD$$
So $$AD = x/\sqrt{2}$$

This method is way too mechanical and prone to errors. Try to use the big picture approach.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 11 Aug 2013
Posts: 28
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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.
Intern  Joined: 04 Jun 2014
Posts: 46
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 60687
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

1
1
lou34 wrote:
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.

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?

Since AC = BC = x, then AC*BC = x^2.

Since $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$, then $$AB * AD =x\sqrt{2}*\frac{x}{\sqrt{2}}=x^2$$.
_________________
Intern  Joined: 16 Mar 2014
Posts: 5
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10021
Location: Pune, India
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

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
_________________
Karishma
Veritas Prep GMAT Instructor Re: In the figure above, is the area of triangular region ABC   [#permalink] 12 Mar 2015, 19:53

Go to page    1   2   3    Next  [ 47 posts ]

Display posts from previous: Sort by

# In the figure above, is the area of triangular region ABC  