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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 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 ? :-)

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

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

Areas of both the triangles is the same



Hello VeritasKarishma

i have two questions :)

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 :)


thanks! :-)
VP
VP
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Joined: 09 Mar 2016
Posts: 1218
Re: In the figure above, is the area of triangular region ABC  [#permalink]

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

Areas of both the triangles is the same



Hello VeritasKarishma

i have two questions :)

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 :)


thanks! :-)
VP
VP
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D
Joined: 09 Mar 2016
Posts: 1218
In the figure above, is the area of triangular region ABC  [#permalink]

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New post 20 Aug 2018, 11:59
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


Bunuel, chetan2u

When \(AD = \frac{AC}{\sqrt{2}}\) doesnt it mean that AC is a hypotenuse :?

Why I am saying so ? Well, because in triangle 45-45-90 when we know only hypotenuse and want to find a leg we divide hypotenuse by \(\sqrt{2}\)

For example hypotenuse is 10 and to find leg i make following step
\(\frac{Hyp}{\sqrt{2}}\) = \(\frac{10}{\sqrt{2}}\) =\(5\sqrt{2}\)

may be you know guys answer to my question :) Probus , CounterSniper pushpitkc would appreciate your reply :)


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

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New post Updated on: 21 Aug 2018, 06:52
1
dave13 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


Bunuel, chetan2u

When \(AD = \frac{AC}{\sqrt{2}}\) doesnt it mean that AC is a hypotenuse :?

Why I am saying so ? Well, because in triangle 45-45-90 when we know only hypotenuse and want to find a leg we divide hypotenuse by \(\sqrt{2}\)

For example hypotenuse is 10 and to find leg i make following step
\(\frac{Hyp}{\sqrt{2}}\) = \(\frac{10}{\sqrt{2}}\) =\(5\sqrt{2}\)

may be you know guys answer to my question :) Probus , CounterSniper pushpitkc would appreciate your reply :)




Hello dave13,

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

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

Areas of both the triangles is the same



Hello VeritasKarishma

i have two questions :)

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?
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Veritas Prep GMAT Instructor

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

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