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# Common Mistakes in Geometry Questions - Exercise Question #2

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Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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Updated on: 07 Aug 2018, 04:00
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Question Stats:

53% (01:43) correct 47% (01:32) wrong based on 544 sessions

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Exercise Question #2 Common Mistakes in Geometry Questions

In the triangle above, what is the measure of Angle CDB?
(1) Angle CAB = Angle ACB = 30°
(2) AB = BC = BD

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.

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Originally posted by EgmatQuantExpert on 22 Nov 2016, 04:32.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:00, edited 3 times in total.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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Updated on: 01 Dec 2016, 00:48
Note: This questions is related to the article on Common Errors in Geometry

Kindly go through the article once, before solving the question or going through the solution.

Official Solution

Find: Value of angle CDB

Working:

Analysing Statement 1:

Given: angle CAB = angle ACB = $$30^o$$.

In triangle ACD –

angle (ACD + CAD + CDA) = $$180^o$$

We know that –

angle ACD = $$90^o$$
angle CAB = $$30^o$$

and CDA can be written as CDB

Therefore,

$$90^o$$ + $$30^o$$ + CDB = $$180^o$$

angle CDB = $$60^o$$.

Hence Statement 1 is sufficient to the answer the question.

Analysing Statement 2:

Given: AB = BC = BD

We can conclude that triangle ABC and triangle CBD are isosceles triangles.

But that will not help us to find the angle CBD, as we cannot find the value of individuals angles of either triangle ABC or triangle CDB.

Hence statement 2 is not sufficient to answer the question.

Since we are getting our answer from the first statement only.

The correct answer option is A

Thanks,
Saquib
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Originally posted by EgmatQuantExpert on 22 Nov 2016, 10:12.
Last edited by EgmatQuantExpert on 01 Dec 2016, 00:48, edited 1 time in total.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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27 Nov 2016, 07:12
Hey Everyone,

The official solution has been posted. Kindly go through it once and feel free to post your doubts, if any.

Thanks,
Saquib
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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10 Aug 2017, 04:00
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EgmatQuantExpert wrote:
Hey Everyone,

The official solution has been posted. Kindly go through it once and feel free to post your doubts, if any.

Thanks,
Saquib
Quant Expert
e-GMAT

EgmatQuantExpert

Can you please explain how statement 2 is insufficient?

Below is my analysis:
Angle CAB + Angle CBA = Angle CBA
Angle BCD + Angle BDC = Angle CBD

Adding both the equations we get:
Angle CAB + Angle CBA + Angle BCD + Angle BDC = 180

as triangle ABC and CBD are isosceles triangle and share a common side so,

4 X Angle BDC = 180
Angle BDC = 45.

Statement 2 is sufficient.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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10 Aug 2017, 16:58
1
EgmatQuantExpert wrote:
Note: This questions is related to the article on Common Errors in Geometry

Kindly go through the article once, before solving the question or going through the solution.

Official Solution

Find: Value of angle CDB

Working:

Analysing Statement 1:

Given: angle CAB = angle ACB = $$30^o$$.

In triangle ACD –

angle (ACD + CAD + CDA) = $$180^o$$

We know that –

angle ACD = $$90^o$$
angle CAB = $$30^o$$

and CDA can be written as CDB

Therefore,

$$90^o$$ + $$30^o$$ + CDB = $$180^o$$

angle CDB = $$60^o$$.

Hence Statement 1 is sufficient to the answer the question.

Analysing Statement 2:

Given: AB = BC = BD

We can conclude that triangle ABC and triangle CBD are isosceles triangles.

But that will not help us to find the angle CBD, as we cannot find the value of individuals angles of either triangle ABC or triangle CDB.

Hence statement 2 is not sufficient to answer the question.

Since we are getting our answer from the first statement only.

The correct answer option is A

Thanks,
Saquib
Quant Expert
e- GMAT

As per statement 2:
AB=BD, thus CB is bisecting AD.

As ACD=90 degress, ACB=BCD=45 degrees.

Statement 2 is also sufficient
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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11 Aug 2017, 00:14
Since BC = BD, which makes CBD 1:1:sqrt 2 right triangle, correct?
So we should be able to know the angle CDB. isn't it?
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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20 Aug 2017, 08:33
Not sure why B is insufficient. According to statement 2, BC is a straight line and perpendicular to AD. It can be inferred that Angle BCD = Angle BDC = 45°

Someone please clarify. Thanks.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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20 Aug 2017, 10:01
I am quite confused regarding statement 2. Initially I thought it to be sufficient.

Experts kindly help , how is it insufficient.
My reasoning is same as posted above by two persons.

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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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20 Aug 2017, 10:27
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Statement 2 is not sufficient here. The solutions above that contend it is sufficient are all making assumptions that you don't know to be true. For example, you don't know that the vertical-looking line bisects the 90 degree angle - you'd need to know that the large triangle was isosceles (AC = CD) to conclude that, and we don't know whether AC = CD is true. Nor do we know that the vertical-looking line is perpendicular to the horizontal-looking line (i.e. we don't know that AD is perpendicular to BC). If you did know either of those things, Statement 2 would be sufficient, but we don't know those things.

You can see visually why Statement 2 is not sufficient. Draw a circle, with center B, and draw a diameter AD. Then let C be any other point on the circumference of the circle. Draw the triangle ADC. We must get a 90 degree angle at C, since we're connecting a diameter to a point on the circle, and every radius is the same length, so BC = BA = BD. By moving the point C around the circumference, you can easily change the size of the two smaller angles in triangle ADC (the angles at A and at D). Now just erase the circle and you have the same situation as is described in this question, when using Statement 2 alone. Since we just saw we can move point C around and change the size of angle CDB, the information can't be sufficient.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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14 Sep 2017, 18:36
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Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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14 Sep 2017, 20:51
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getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?

Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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16 Jan 2018, 14:43
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?

Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?

It seems that I am assuming wrongly that angle ACB = angle BCD = angle CDB since AB = BD = BC and each of those angles is opposite each respective side. Since ACB + BCD = 90 degrees and ACB = BCD, then ACB and BCD = 45 degree each. Thus CDB = 45 degrees.

Any chance you could help me identify where my thinking is flawed? A few months later and I still can't figure it out. Thanks a ton.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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16 Jan 2018, 18:59
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?

Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?

Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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16 Jan 2018, 19:59
3
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?

Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?

Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?

Useful property: The median on the hypotenuse of a right triangle always equals to one-half the hypotenuse.

AB = BC = BD means that CB is the median (B is the midpoint of AD). Because of the property above, AB = BC = BD is true for any right angled triangle with median CB. So, CB may or may not be perpendicular to AD.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2  [#permalink]

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27 Oct 2018, 20:14
EgmatQuantExpert wrote:

In the triangle above, what is the measure of Angle CDB?
(1) Angle CAB = Angle ACB = 30°
(2) AB = BC = BD

$$? = \angle CDB = x$$

(All angles are measured in degrees.)

(1) Using the exterior angle property (to justify 30+30 in red), triangle BCD is equilateral (hence x=60). Sufficient!

(2) The constructions shown below are viable and present two different possible values for x. Insufficient!

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Common Mistakes in Geometry Questions - Exercise Question #2 &nbs [#permalink] 27 Oct 2018, 20:14
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# Common Mistakes in Geometry Questions - Exercise Question #2

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