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# Mastering Important Concepts in Triangles - Exercise Question #2

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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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43% (02:01) correct 57% (02:08) wrong based on 253 sessions

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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

A. 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:

Detailed solution will be posted soon.

Thanks,
Saquib
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Originally posted by EgmatQuantExpert on 28 Nov 2016, 01:15.
Last edited by EgmatQuantExpert on 02 Dec 2016, 09:03, edited 8 times in total.
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Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]

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Updated on: 07 Dec 2016, 04:01
3
Let's look at the detailed solution of the above problem

Steps 1 & 2: Understand Question and Draw Inferences
Given a quadrilateral ABCD with right angles at B and D.
We need to find if AD > BC is true.

Step 3: Analyze Statement 1 independently
Given 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 independently
Given 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,$$BC^2 > AD^2$$
• $$BC > AD$$

Therefore, statement 2 is sufficient to arrive at a unique answer.

Hence the correct Answer is B.

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Originally posted by EgmatQuantExpert on 28 Nov 2016, 01:16.
Last edited by EgmatQuantExpert on 07 Dec 2016, 04:01, edited 2 times in total.
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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
gracie90 wrote:
Where is the actual question?

Hey,

The question has been updated, please check.

Thanks,
Saquib
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e-GMAT
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Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD
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Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line

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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]

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02 Dec 2016, 00:12
3
1
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.

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Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]

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Updated on: 05 Dec 2016, 07:27
1
1
Let's look at the detailed solution of the above problem

Steps 1 & 2: Understand Question and Draw Inferences
Given a quadrilateral ABCD with right angles at B and D.
We need to find if AD > BC is true.

Step 3: Analyze Statement 1 independently
Given 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 independently
Given 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,$$BC^2 > AD^2$$
• $$BC > AD$$

Therefore, statement 2 is sufficient to arrive at a unique answer.

Hence the correct Answer is B

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Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line

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Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

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Originally posted by EgmatQuantExpert on 05 Dec 2016, 07:14.
Last edited by EgmatQuantExpert on 05 Dec 2016, 07:27, edited 1 time in total.
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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 45-45-90 rule).

Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.

Please correct me. Where I am loosing the track?

Thanks.
Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1393
Location: Viet Nam
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 45-45-90 rule).

Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.

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 45-45-90 rule).

Now AC= sqrt of 2 * x and Angle ADC=90 degree, AD=DC must be x.

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 $$45^o-45^o-90^o$$ triangle, and hence you wrote AD=DC, which is not the case. From Statement (1), we can only conclude that triangle ABC is a $$45^o-45^o-90^o$$ 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 questions

Regards,
Saquib
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Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

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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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