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Mastering Important Concepts in Triangles - Exercise Question #2

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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
Where is the actual question?
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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
gracie90 wrote:
Where is the actual question?

Hey,

The question has been updated, please check.

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

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

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

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 $$45^o-45^o-90^o$$ 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

Originally posted by EgmatQuantExpert on 15 Dec 2016, 23:57.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:20, edited 1 time in total.
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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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
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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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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.
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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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Re: Mastering Important Concepts in Triangles - Exercise Question #2 [#permalink]
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