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We cannot simplify or reduce further. Insufficient.

(1) and (2)

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

\(AB = AD*\sqrt{2}?\)

Sufficient. Answer is C.

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!

Re: In the figure above, is the area of triangular region ABC [#permalink]
23 Oct 2012, 21:39

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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 17361 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:

(1) (AC)^2=2(AD)^2 \(AD = AC/\sqrt{2}\) 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 17475 times ]

(2) ∆ABC is isosceles. We have no idea about the length of AD so insufficient.

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 _________________

Re: In the figure above, is the area of triangular region ABC [#permalink]
24 Oct 2012, 01:04

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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 ?

(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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In the figure above, is the area of triangular region ABC [#permalink]
05 Jun 2013, 17:41

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

Re: In the figure above, is the area of triangular region ABC [#permalink]
05 Jun 2013, 23:07

1

This post received KUDOS

Expert's post

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

There's your error. \(AD = AC/\sqrt{2}\) So AD < AC (not AD > AC) _________________

Re: In the figure above, is the area of triangular region ABC [#permalink]
18 Sep 2013, 16:14

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 \(AD=\frac{x}{\sqrt{2}}\)?

Re: In the figure above, is the area of triangular region ABC [#permalink]
18 Sep 2013, 17:49

Expert's post

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 ?

(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 \(AD=\frac{x}{\sqrt{2}}\)?

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. _________________

Re: In the figure above, is the area of triangular region ABC [#permalink]
18 Nov 2013, 20:37

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.

Re: In the figure above, is the area of triangular region ABC [#permalink]
12 Mar 2015, 07:58

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.

Re: In the figure above, is the area of triangular region ABC [#permalink]
12 Mar 2015, 18:53

Expert's post

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.

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