Mastering Important Concepts in Triangles - Exercise Question #2

Question

Mastering Important Concepts Tested By GMAT in Triangles - Exercise Question #2

In the figure above - Angle ABC = Angle ADC = $90^o$. Is AD > BC?

Answer Choices

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
e-GMAT

Show Answer

Official Answer

B

Thanks, Saquib Quant Expert e-GMAT https://e-gmat.com/blogs/wp-content/uploads/2016/12/Geometry-Ability-Quiz-4-e1480581478957.png

EgmatQuantExpert · Accepted Answer

Let's look at the detailed solution of the above problem

https://e-gmat.com/blogs/wp-content/uploads/2016/11/Q370-e1480320572599.png

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.

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

broall · Answer

(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

gracie90 · Answer

Where is the actual question?

EgmatQuantExpert · Answer

Hey,

The question has been updated, please check.

Thanks,
Saquib
Quant Expert
e-GMAT

EgmatQuantExpert · Answer

Let's look at the detailed solution of the above problem

https://e-gmat.com/blogs/wp-content/uploads/2016/11/Q370-e1480320572599.png

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

goalMBA90 · Answer

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.

So AD=BC. We can surely tell that No, AD>BC.

Please correct me. Where I am loosing the track?

Thanks.

broall · Answer

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.

EgmatQuantExpert · Answer

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

7sharma · Answer

why CD-AB=BC-AD ..

EatMyDosa · Answer

S-1: Gives BC = AB. Since s-1 gives no relation between BC and AD, it is NOT SUFFICIENT

S-2: In triangle ABC, angle BAC = 90 - angle ACB ----- (1)

In triangle ADC, angle ACD = 90 - angle CAD ----- (2)

Since per s-2, CD > AB, it implies that

angle CAD > angle ACB -------- (3)

From equations (1), (2), and (3) combined, we have

angle ACD < angle BAC, therefore

AD < BC SUFFICIENT

WheatyPie · Answer

I solved this by visualizing. Imagine that AC is the diameter of a circle. Then, since angles ADC and ABC are both 90, points D and B would both lie on the circle.

Statement 1 tells us that AB and BC are congruent, but by moving point D around the circle I can make side AD bigger than or smaller than side BC. Not sufficient.

Statement 2 tells us that CD > AB ... imagined in a circle, it's fairly clear visually that this forces BC to be bigger than AD. Sufficient.

shauryahanda · Answer

I would like to point out something here

No where in the question it is mentioned that the figure is not to scale so considering the figure to be close to what it is

We can infer from st1

1) Let AC = x, BC = x/sqrt(2)

Now we know <ACD < 45 otherwise angle C = 90 and DC would be parallel to AB which is not the case based on the figure

Hence AD = xsin(theta) where theta < 45 
HENCE AD < x/sqrt(2)
therefore AD < BC

and hence A is sufficient

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Mastering Important Concepts in Triangles - Exercise Question #2

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Mastering Important Concepts in Triangles - Exercise Question #2

Prep Toolkit

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