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Re: In the figure above, is the area of triangular region ABC
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23 Oct 2012, 21:39
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Expert Reply
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 116412 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 116309 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 _________________
Karishma GMAT Instructor at Angles and Arguments https://anglesandarguments.com/
Re: In the figure above, is the area of triangular region ABC
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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.
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]
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
2
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Expert Reply
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. _________________
Karishma GMAT Instructor at Angles and Arguments https://anglesandarguments.com/
Re: In the figure above, is the area of triangular region ABC
[#permalink]
13 Apr 2015, 01:41
1
Kudos
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
Re: In the figure above, is the area of triangular region ABC
[#permalink]
13 Apr 2015, 17:45
Expert Reply
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.
Re: In the figure above, is the area of triangular region ABC
[#permalink]
14 Apr 2015, 00:31
1
Kudos
EMPOWERgmatRichC wrote:
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.
Re: In the figure above, is the area of triangular region ABC
[#permalink]
14 Apr 2015, 21:35
Expert Reply
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.
In the figure above, is the area of triangular region ABC
[#permalink]
13 Aug 2020, 10:33
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.
Re: In the figure above, is the area of triangular region ABC
[#permalink]
13 Aug 2020, 19:11
Expert Reply
Avinashkr29 wrote:
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.
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
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.
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Answer: Option C
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