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Re: In the figure above, is the area of triangular region ABC
[#permalink]
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]
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]
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
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Re: In the figure above, is the area of triangular region ABC
[#permalink]
21 Aug 2018, 06:46
1
Kudos
Expert Reply
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
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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.
Re: In the figure above, is the area of triangular region ABC
[#permalink]
16 Nov 2020, 01:44
I think this question can be solved without math, but through knowledge of concepts. See attached image showing scratchpad work.
We're trying to see whether we have enough information to determine the areas of both triangles or get an idea of the relationship of the areas of the two triangles. We can see that AB is the hypotenuse of triangle ABC, so we know that one leg of triangle ADB is related to triangle ABC - this means that if we have information on three legs (AD, AC and CB), then we have enough information to answer the question.
1) gives us a relationship between AD and AC, we just need to know about CB --> Insufficient 2) gives us a relationship between AC and CB, we just need to know about AC --> Insufficient
Combining 1) and 2) statements together gives us information on all three legs --> Sufficient
Re: In the figure above, is the area of triangular region ABC
[#permalink]
02 Feb 2021, 01:58
I have a question for the experts. I'm new to GMAT Club. Please help!!
I was able to deduce from the given information (AB^2=AC^2+CB^2) and the information in statement one (AC^2 = 2*AD^2) that AC = CB (making ACB isosceles). This would make the statements 1 and 2 redundant, and therefore the answer would be "D." Did I do something incorrectly in the math below? Thank you for your advice in advance!
Statement 1: AC^2 = 2*AD^2 --> (1/2)*AC^2=AD^2 --> used this equation to substitute AD^2 in the following equation: AC^2*BC^2=AB^2*AD^2 AC^2*BC^2=AB^2*(1/2)*AC^2 BC^2=AB^2*(1/2) BC^2=[AC^2+CB^2]*(1/2) [taken from the pythagorean theorem above] BC^2=AC^2*(1/2)+CB^2*(1/2) BC^2*(1/2)=AC^2*(1/2) therefore BC=AC
Re: In the figure above, is the area of triangular region ABC
[#permalink]
25 Feb 2021, 08:48
(1) Clearly insufficient since the base of ABC is unknown. Both areas are dependent on the value of the base of ABC. (2) Clearly unknown since the side of AD is not known.
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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