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 Your Progress

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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

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

Show Tags

19 Aug 2018, 06:18

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.

hi there

how did you get from here \(\sqrt{2}AD\cdot{BC}=AB\cdot{AD}\), this \(\sqrt{2}BC=AB\)

you write " This means that triangle \(ABC\) should necessarily be isosceles, which we don't know. isnt here a word NOT omitted here

shouldnt it be like this "This means that triangle \(ABC\) should NOT necessarily be isosceles, which we don't know"

and how can i looking at this \(\sqrt{2}BC=AB\) assess if trinagle is an iscoleses or not or any kind ? what is your thought process ?

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

Show Tags

19 Aug 2018, 06:40

VeritasKarishma wrote:

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

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

(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 = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\)

Question # 1. how from this (1) \((AC)^2=2(AD)^2\) you got this \(AD = AC/\sqrt{2}\)

here is how i did it but something i did wrong please advise.

\((AC)^2=2(AD)^2\)

re-arrange

\(2(AD)^2 = (AC)^2\)

\((AD)^2 = \frac{(AC)^2}{2}\) dividing by 2

\(\sqrt{(AD)^2}= \sqrt{\frac{(AC)^2}{2}}\) now squaring both sides

\(AD = \frac{AC}{2}\)

Question # 2.

Area of DAB = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\) which side of triangle does a highliged part represent and how you got the highlighted part

well thats all i wanted to ask hope you have a time to reply

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

Show Tags

19 Aug 2018, 06:41

VeritasKarishma wrote:

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

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

(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 = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\)

Question # 1. how from this (1) \((AC)^2=2(AD)^2\) you got this \(AD = AC/\sqrt{2}\)

here is how i did it but something i did wrong please advise.

\((AC)^2=2(AD)^2\)

re-arrange

\(2(AD)^2 = (AC)^2\)

\((AD)^2 = \frac{(AC)^2}{2}\) dividing by 2

\(\sqrt{(AD)^2}= \sqrt{\frac{(AC)^2}{2}}\) now squaring both sides

\(AD = \frac{AC}{2}\)

Question # 2.

Area of DAB = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\) which side of triangle does a highliged part represent and how you got the highlighted part

well thats all i wanted to ask hope you have a time to reply

I will explain the solution Draw two right triangles on sheet of paper ACB right angled at C, and DAB right angled at A

Now we are asked if the area of two triangles are same.

How can we say if the area of two triangles are same 1. The two triangles are congruent. that is each side of one triangle is congruent with corresponding side of the other triangle. 2 We know the measurements of the base and height of two right triangles and we calculate the area for each

So we are asked if area of \(\triangle\) ACB = area of \(\triangle\) DAB \(\frac {1}{2}\)* AC*BC= \(\frac {1}{2}\)AD*AB more so if AC*BC= AD*AB -------- Say Base ondition

Now Stmt 1: \({AC}^2\)= 2\({AD}^2\) So taking Square root on both sides AC=\(\sqrt{2}\) * AD

Now if we substitute this in our base condition then

\(\sqrt{2}\) * AD * BC= AD*AB AD cancels out \(\sqrt{2}\) * BC = AB

Now if you Look at right \(\triangle\) ACB we have \(\sqrt{2}\) * BC = AB . Now in 45-45 -90 Traingle i will give a simple rule which i keep for my understanding 1. IF X : the \(\triangle\) is right angled and isosceles , then the angles of Triangle are 45-45-90 Then Y: their sides are in the ratio of 1:1:\(\sqrt{2}\)

SO X leads to Y

But we have have found out that \(\sqrt{2}\) * BC = AB, which is Y . But does this mean that X must be there. NO AC could be any other measurement. or the angles if the triangle could be anything except 45-45-90

Now STMT 2: However, in absence of any information triangle DAB lets see how STMT 2 would it help if we knew that ACB is isosceles

area = \(\frac {1}{2}\) AC*BC or area =\(\frac {1}{2}\) \(BC^{2}\) becuase AC=BC

SO area of Triangle ACB =\(\frac {1}{2}\) \(BC^{2}\) SAY EQUATION (1) But we don't know anything about the other Triangle DAB so we can't conclude anyhting

Now If we combine both information then from Triangle DAB we have

area = 1/2 AD* AB \(\frac {1}{2}\)* \(\frac {AC}{\sqrt{2}}\), * \(\sqrt{2}\) * BC Becuase AB= \(\sqrt{2}\) * BC,& AD= \(\frac {AC}{\sqrt{2}}\)

So we have \(\frac {1}{2}\)* AC*BC and since we know that AC= BC = \(\frac {1}{2}\)*\(BC^{2}\) SAY equation (2)

equation 1 = equation 2 hence both are required

Probus

Originally posted by Probus on 21 Aug 2018, 06:08.
Last edited by Probus on 21 Aug 2018, 06:52, edited 2 times in total.

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

Show Tags

21 Aug 2018, 06:46

1

dave13 wrote:

VeritasKarishma wrote:

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

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

(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 = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\)

Question # 1. how from this (1) \((AC)^2=2(AD)^2\) you got this \(AD = AC/\sqrt{2}\)

here is how i did it but something i did wrong please advise.

\((AC)^2=2(AD)^2\)

re-arrange

\(2(AD)^2 = (AC)^2\)

\((AD)^2 = \frac{(AC)^2}{2}\) dividing by 2

\(\sqrt{(AD)^2}= \sqrt{\frac{(AC)^2}{2}}\) now squaring both sides

\(AD = \frac{AC}{2}\)- Incorrect

Note that \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\)

So \(\sqrt{\frac{(AC)^2}{2}} = \frac{AC}{\sqrt{2}}\)

Quote:

Question # 2.

Area of DAB = \(1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2\) which side of triangle does a highliged part represent and how you got the highlighted part

Using both statements, we know that

\(1:1:\sqrt{2} = AC:BC : AB\) and \(AD = AC/\sqrt{2}\)

So, if AC = 1, BC is also 1, \(AB =\sqrt{2}\) and \(AD = 1/\sqrt{2}\) These are the values we put in for AD and AB.

Does this help?
_________________

Karishma Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

gmatclubot

Re: In the figure above, is the area of triangular region ABC &nbs
[#permalink]
21 Aug 2018, 06:46