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

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Intern
Joined: 17 Sep 2011
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In the figure above, is the area of triangular region ABC  [#permalink]

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10 Jun 2012, 00:28
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62% (02:25) correct 38% (02:22) wrong based on 2511 sessions

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

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23 Oct 2012, 22:39
39
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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
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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24 Oct 2012, 02:04
40
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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.

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

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10 Jun 2012, 01:13
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.

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

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10 Jun 2012, 02:00
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!
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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23 Oct 2012, 11:36
does anyone have any other alternative method to solve?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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13 Mar 2013, 23:53
Nice and succinct explanation Eva...
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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05 Jun 2013, 18:41
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
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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06 Jun 2013, 00:07
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

$$AD = AC/\sqrt{2}$$
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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06 Jun 2013, 17:55
Ah, I get it. Thank you for pointing my error.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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24 Aug 2013, 03:49
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?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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24 Aug 2013, 03:54
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.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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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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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 ?

(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 $$AD=\frac{x}{\sqrt{2}}$$?
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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18 Sep 2013, 18:49
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.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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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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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.

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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12 Aug 2014, 06:49
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 $$AB * AD =x\sqrt{2}*\frac{x}{\sqrt{2}}=x^2$$.
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Re: In the figure above, is the area of triangular region ABC  [#permalink]

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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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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] 12 Mar 2015, 19:53

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