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

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.

[strike]AD[/strike]
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.

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

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

Thanks a lot!

does anyone have any other alternative method to solve?

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.

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 \(AB * AD =x\sqrt{2}*\frac{x}{\sqrt{2}}=x^2\).

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

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

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

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

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.

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

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

Let's start by labeling the side lengths as follows:


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:


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:


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:


Statement 2 tells us that AC = CB, which means we can rewrite the length of CB as follows:


Now let's focus on ∆ABC
Applying the Pythagorean theorem we can write: [(√2)y]² + [(√2)y]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:

This means the area of ∆ABC = (1/2)[(√2)y][(√2)y] = 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

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

Answer: Option C

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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