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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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Updated on: 02 Dec 2016, 09:03
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Mastering Important Concepts Tested By GMAT in Triangles  Exercise Question #2
In the figure above  Angle ABC = Angle ADC = \(90^o\). Is AD > BC?
(1) Angle CAB = Angle BCA (2) CD > AB
Answer ChoicesA. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient. Key concepts on Triangles are covered in details in the following post: Mastering Important Concepts Tested By GMAT in Triangle  I Detailed solution will be posted soon. Thanks, Saquib Quant Expert eGMAT
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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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Updated on: 07 Dec 2016, 04:01
Let's look at the detailed solution of the above problem Steps 1 & 2: Understand Question and Draw InferencesGiven a quadrilateral ABCD with right angles at B and D. We need to find if AD > BC is true. Step 3: Analyze Statement 1 independentlyGiven that \(Angle CAB = Angle BCA\) Therefore, in \(ΔABC\), \(Angle CAB = Angle BCA = 45^o\) Therefore, \(BC = AB\) This doesn’t give us any relation between AD and BC. So statement 1 is not sufficient to arrive at a unique answer.Step 4: Analyze Statement 2 independentlyGiven that \(CD > AB\). In the right angled triangle ABC, we have \(AC^2 = AB^2 + BC^2\) Similarly, in the right angled triangle ADC, we have \(AC^2 = AD^2 + CD^2\) Therefore, \(AD^2 + CD^2 = AB^2 + BC^2\)……………. (I) Given that CD > AB • \(CD^2 > AB^2\) In other words, a term in the LHS of (I) is greater than another term in the RHS of (I) So for (I) to be true (the equality to hold true), the remaining terms on either sides of LHS and RHS should compensate for the imbalance created by \(CD^2 and AB^2\). From (I), \(CD^2 AB^2 = BC^2 AD^2\) Let’s say \(CD^2 – AB^2 = k\)(a positive number since \(CD^2 > AB^2\)) Then \(BC^2 – AD^2 = k\) (the same positive number). Therefore,\(BC^2 > AD^2\) Therefore, statement 2 is sufficient to arrive at a unique answer.
Hence the correct Answer is B.
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Re: Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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29 Nov 2016, 12:37
Where is the actual question?



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Re: Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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29 Nov 2016, 13:29



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Re: Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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02 Dec 2016, 00:12
EgmatQuantExpert wrote: In the figure above  Angle ABC = Angle ADC = \(90^o\). Is AD > BC? (1) Angle CAB = Angle BCA (2) CD > AB
(1) \(\widehat{CAB}=\widehat{BCA}\) so \(\Delta ABC\) is an isosceles right triangle \(\implies AB=BC\). However, we can't know whether \(AD>BC\) or not. Insufficient. (2) We have \(AC^2=AB^2+BC^2=AD^2 + CD^2\). Since \(CD>AB \implies CD^2>AB^2 \implies AD^2 < BC^2 \implies AD<BC\). Sufficient. The answer is B
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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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Updated on: 05 Dec 2016, 07:27
Let's look at the detailed solution of the above problem Steps 1 & 2: Understand Question and Draw InferencesGiven a quadrilateral ABCD with right angles at B and D. We need to find if AD > BC is true. Step 3: Analyze Statement 1 independentlyGiven that \(Angle CAB = Angle BCA\) Therefore, in \(ΔABC\), \(Angle CAB = Angle BCA = 45^o\) Therefore, \(BC = AB\) This doesn’t give us any relation between AD and BC. So statement 1 is not sufficient to arrive at a unique answer.Step 4: Analyze Statement 2 independentlyGiven that \(CD > AB\). In the right angled triangle ABC, we have \(AC^2 = AB^2 + BC^2\) Similarly, in the right angled triangle ADC, we have \(AC^2 = AD^2 + CD^2\) Therefore, \(AD^2 + CD^2 = AB^2 + BC^2\)……………. (I) Given that CD > AB • \(CD^2 > AB^2\) In other words, a term in the LHS of (I) is greater than another term in the RHS of (I) So for (I) to be true (the equality to hold true), the remaining terms on either sides of LHS and RHS should compensate for the imbalance created by \(CD^2 and AB^2\). From (I), \(CD^2 AB^2 = BC^2 AD^2\) Let’s say \(CD^2 – AB^2 = k\)(a positive number since \(CD^2 > AB^2\)) Then \(BC^2 – AD^2 = k\) (the same positive number). Therefore,\(BC^2 > AD^2\) Therefore, statement 2 is sufficient to arrive at a unique answer.
Hence the correct Answer is B
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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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13 Dec 2016, 12:37
Why statement 1 is not sufficient? Bcoz Angle ABC = Angle ADC = 90 degree given,
So AB=BC=x and AC=sqrt of 2 * x (as per 454590 rule).
Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.
So AD=BC. We can surely tell that No, AD>BC.
Please correct me. Where I am loosing the track?
Thanks.



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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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13 Dec 2016, 18:18
goalMBA90 wrote: Why statement 1 is not sufficient? Bcoz Angle ABC = Angle ADC = 90 degree given,
So AB=BC=x and AC=sqrt of 2 * x (as per 454590 rule).
Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.
So AD=BC. We can surely tell that No, AD>BC.
Please correct me. Where I am loosing the track?
Thanks. In statement (1), there is no such condition as the highlighted. Since we can't know AD=DC, AD>DC or AD<DC, then (1) is insufficient.
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Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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15 Dec 2016, 23:57
goalMBA90 wrote: Why statement 1 is not sufficient? Bcoz Angle ABC = Angle ADC = 90 degree given,
So AB=BC=x and AC=sqrt of 2 * x (as per 454590 rule).
Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.
So AD=BC. We can surely tell that No, AD>BC.
Please correct me. Where I am loosing the track?
Thanks. Dear Student, I believe that while analyzing Statement (1), by mistake you assumed that triangle ADC is also a \(45^o45^o90^o\) triangle, and hence you wrote AD=DC, which is not the case. From Statement (1), we can only conclude that triangle ABC is a \(45^o45^o90^o\) triangle, we cannot infer the same about triangle ADC. This is one of the common mistake done by people while solving Geometry questions. Try not to assume any information, which is not given in the question. We recommend that you go through the following article to avoid Common mistakes done by students while solving geometry questions. Common Mistakes in Geometry questionsRegards, Saquib
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Re: Mastering Important Concepts in Triangles  Exercise Question #2 [#permalink]
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10 Apr 2018, 08:11
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