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

Question

https://gmatclub.com/forum/download/file.php?id=17552 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 OG13DS79v2.png

KarishmaB · Accepted Answer

You can solve sufficiency questions of geometry by drawing some diagrams too. 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. 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 This question is discussed in this blog post in detail: https://anaprep.com/geometry-drawing-extreme-diagrams/

EvaJager · Answer

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.

geometric · Answer

I started with
Area = .5bh
Area(ABC) = .5*AC*CB
Area(DBA) = .5*AD*AB

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

1) (AC)^2 = 2(AD)^2

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.

AD 
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.

:. Answer is C.

jfk · Answer

Thanks, vandygrad11!

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

Start with rephrasing the question:

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

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

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 :lol:

Hope it helps others. Please let me know if I made any mistake!

Thanks a lot!

Hope it helps others. Please let me know if I made any mistake! Thanks a lot!

teal · Answer

does anyone have any other alternative method to solve?

russ9 · Answer

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}} ?

KarishmaB · Answer

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.

lou34 · Answer

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?

Bunuel · Answer

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 .

erikvm · Answer

Is a valid solution to this to plug in numbers instead of X to avoid painful calculations?

I simply assigned "AC" to 6 --> (AC)^2 = 2(AD)^2 --> 36 = 2(AD)^2. --> 18 =(AD)^2. AD =  3\sqrt{2}

Jumping straight to (1)+(2): So, they are Isoceles: 45-45-90 ratio aka  6-6-6\sqrt{2} . Which means Area of ABC = 36/2 = 18

Area of DBA =  6\sqrt{2} * 3\sqrt{2}  / 2 = 18

I mean... as long as Im conistent with keeping "AC" to 6, it shouldn't really matter since i'm just expressing AC in a different way, right? I just find it way easier to work with actual numbers than with letters

EMPOWERgmatRichC · Answer

Hi erikvm,

Yes, TESTing VALUES on this question is a great way to deal with a situation that is really 'technical.' As long as you're labeling your work and being thorough, you'll find that this approach works on LOTS of Quant questions.

GMAT assassins aren't born, they're made,
Rich

erikvm · Answer

Thanks Rich. I am a big fan of testing values, but got a bit sceptical when it comes to DS questions.

EMPOWERgmatRichC · Answer

Hi erikvm,

TESTing VALUES is a useful approach on most DS questions; you just have to be sure to adjust the tactic a bit. On a PS question, you're (typically) looking for the one answer that matches your values. On a DS question, you're TESTing multiple VALUES to see if a pattern emerges (or if you can prove that a pattern does NOT exist). Since that strategy works in so many different ways on Test Day, it's a great approach to practice throughout your studies.

GMAT assassins aren't born, they're made,
Rich

Avinashkr29 · Answer

Actually I feel Option A should be correct as it is sufficient.
ac^2=2(ad)^2 this implies ac=(root 2)(ad). meaning, ad is bigger than ac.
In any right angles triangle, hypotenuse is always the longest side on triange. so, ab is greater than ac and bc. 
so, area of ABC = 1/2 * AC * BC
where as area of ADB = 1/2 * AB (which is greater than AC and BC) * AD (which is greater than AC).
so area of ADB should be greater than area of ABC.

EMPOWERgmatRichC · Answer

Hi Avinashkr29,

There's one small error in your approach that impacts the entire explanation. While you are correct that...

ac^2=2(ad)^2 this implies ac=(root 2)(ad)

...since you are multiplying (ad) by a number that is GREATER than 1 (re: root 2), that means that (ac) > (ad). Your overall explanation assumes that (ad) > (ac), which is not correct. 
 
GMAT assassins aren't born, they're made,
Rich

BrentGMATPrepNow · Answer

Target question :    Does area of ∆ABC equal the area of ∆DBA?

Let's start by labeling the side lengths as follows:
 https://i.imgur.com/XGAV4GG.png

Statement 1 : (AC)²=2(AD)²  
In other words, w² = 2y²
If we take the square root of both sides, we get: w = (√2)y
So, in the diagram, let's replace w with (√2)y to get: 
 https://i.imgur.com/Ys4GayJ.png

At this point, the relationship between sides AC and AD is "locked" in, but that isn't enough to lock in the answer to the   target question.   
Notice that we can create diagrams that satisfy statement 1, yet yield different answers to the target question. 
Consider these two diagrams: 
 https://i.imgur.com/uVFCxpt.png

For the diagram on the left side, the answer to the target question is  YES, the area of ∆ABC equals the area of ∆DBA 
For the diagram on the right side, the answer to the target question is  NOT, the area of ∆ABC does not equal the area of ∆DBA

Since we can’t answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : ∆ABC is isosceles 
This means AC = CB

IMPORTANT :  For geometry Data Sufficiency questions, we’re typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in the video below.

Notice that statement 2 "locks" in the relationship between sides AC and CB, but we can still mentally grab point D and change the area of ∆DBA without affecting the area of ∆ABC.
In other words, the answer to the target question can be  YES or NO 
Since we can’t answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
From Statement 1, we were able to rewrite the length of AC as follows:
 https://i.imgur.com/Ys4GayJ.png

Statement 2 tells us that AC = CB, which means we can rewrite the length of CB as follows:
 https://i.imgur.com/axaAg4x.png

Now let's focus on ∆ABC
Applying the Pythagorean theorem we can write:  ² +  ² = x²
Simplify to get: 2y² + 2y² = x²
Simplify: 4y² = x²
Take the square root of both sides to get: 2y = x

So let's replace x with 2y to get:  https://i.imgur.com/H1G2str.png

This means the area of ∆ABC = (1/2)   = y²
And the area of ∆DBA = (1/2)(y)(2y) = y²

So, the answer to the target question is  YES, the area of ∆ABC equals the area of ∆DBA 
Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=wetqSGMdY7Y

ScottTargetTestPrep · Answer

Solution:

Question Stem Analysis:

We need to determine whether the two triangles, ABC and DBA, have the same area. Notice that since both triangles are right triangles, the area of triangle ABC = ½ x AC x BC, and the area of triangle DBA = ½ x AD x AB. Finally, notice that AB is the hypotenuse of triangle ABC; therefore, AC^2 + BC^2 = AB^2.

Statement One Alone:

We see that AC = AD√2. However, without knowing BC, we can’t determine the area of either triangle. Statement one alone is not sufficient.

Statement Two Alone:

We see that triangle ABC is an isosceles right triangle, i.e., a 45-45-90 triangle, so that AC = BC and AB = AC√2. However, without knowing AD, we can’t determine the area of triangle DBA and how it is compared to that of triangle ABC. Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, we can let AC = s = BC, so the area of triangle ABC is ½ s^2. From statement one, we have AC = AD√2. Therefore, AD = AC/√2 = s/√2. From statement two, we have AB = AC√2. Therefore, AB = s√2. Since the area of triangle DBA = ½ x AD x AB, the area of triangle DBA = ½ x s/√2 x s√2 = ½ s^2. We see that the two triangles have the same area. Both statements together are sufficient.

Answer: C

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=gdsaA8Jw1Bk

Asha7 · Answer

As per my analysis, St1 is alone sufficient.

ST1: AC^2=2*AD^2---eq2

QS analysis: we need to find AC*CB=AD*AB & squaring on both sides gives us AC^2*CB^2=AD^2*AB^2 ----eq0

so we already know that AB^2=AC^2+BC^2 and if we sub. this AB^2 in eq0 we get

AC^2*CB^2=AD^2(AC^2+BC^2)---eq1 => and from st1 we already know that AC^2=2*AD^2 ---eq2, now sub. eq2 into eq1 we get

2*AD^2*CB^2=2*AD^4+AD^2*BC^2, now moving AD^2*BC^2 in the RHS to LHS we get

AD^2*CB^2=2*AD^2*AD^2 => so CB^2=2*AD^2 i.e AC^2=2*AD^2=CB^2 i.e AC=BC i.e the triangle is a isosceles triangle as AC=BC---eq4

Now sub. eq4 into eq1 we get AC^4=AD^2(2*AC^2), we already are given from st1 that AC^2=2*AD^2

W.r.t LHS AC^4=(AC^2)^2= (2*AD^2)^2=4*AD^4 ----eq5

Now w.r.t RHS we get AD^2*2*AC^2 = 4AD^4 ----eq6

Hence we get LHS=RHS -- This is the reason i feel st1 alone is sufficient.

Please correct me if i am wrong

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards