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M01 #21

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M01 #21 [#permalink] New post 05 Feb 2009, 15:25
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

63% (01:55) correct 37% (01:06) wrong based on 190 sessions
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B

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Re: M01 #21 [#permalink] New post 21 Aug 2009, 09:42
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Hi crejoc,

From stmt1:
ACB = 2x, ADC=x

ACD = 180-2x
Now in triangle ACD, ACD + ADC + DAC = 180
=> 180-2x+x+DAC = 180
=> DAC = x = ADC
OR, CAD=CDA
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Re: M01 #21 [#permalink] New post 03 Nov 2010, 08:02
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1) Let's say angle ADC is 30 making angle ACB = 60. That would make triangle ABC a 30, 60, 90 triangle so the answer would be NO.
However, ADC can be 22.5 or any angle for that matter but let's say it's 22.5 which makes angle ACB = 45. That would make triangle ABC an isosceles right triangle with the answer YES. This is NOT sufficient.

2) If angle ACB is twice as large as angle CAB we can write it as ACB = 2CAB. We know that angle ABC = 90 degrees so we use the following equations.

ACB = 2CAB

ABC + ACB + CAB = 180 (let's put 90 for ABC since we know that angle)
90 + ACB + CAB = 180 (we can plug in 2CAB in place of ACB from here to get CAB's angle)

90 + 2CAB + CAB = 180
90 + 3CAB = 180
Solving for CAB we get 30 degrees

Triangle ABC can ONLY be a 30, 60, 90 triangle and the answer will always be NO so selection (2) is sufficient.
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Re: M01 #21 [#permalink] New post 25 Mar 2012, 11:09
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prakarp wrote:
I have a doubt . from crejoc's reply I get that angle CAD= angle ADC . Now considering the triangle ADC the exterior angle ACB = sum of angles( CAD+ CDA) . So angle ACB > angle ADC . Now taking the rule angle opposite to greater side is greater and comparing the angles ACB and ADC we have AB> AC . Hence ABC is not isosceles.

So should n't the answer be D instead of B.... please help...


AB is a leg in a right triangle ABC and it cannot be more than the hypotenuse AC of the same right triangle.

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In a triangle ABD, B is the right angle. Point C is on BD. If you draw a line from A to C, then is triangle ABC an isosceles triangle?

Notice that since ABC is a right triangle then the question basically asks whether angle ACB is equal to angle CAB, because in this case AB would be equal to BC, thus triangle ABC would be an isosceles triangle.

(1) Angle ACB is twice as large as angle ADC --> it certainly possible angle ACB to be equal to angle CAB (45 degrees) and angle ADC to be 22.5 degrees but it's also possible angle ACB NOT to be equal to angle CAB, for example if angle ACB is 60 degrees and angle ADC is 30 degrees. Not sufficient.

(2) Angle ACB is twice as large as angle CAB --> directly says that angle ACB is NOT equal to angle CAB, hence triangle ABC is NOT isosceles. Sufficient.

Answer: B.
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Re: M1 - 21 [#permalink] New post 05 Feb 2009, 19:26
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botirvoy wrote:
I'd appreciate some shortcuts to solve this q.


B.

1: angle D could be 22.5 or anything else.. nsf..
2: If angles ACB is twice large as angle CAB, then traingle ABC is not an isosceles.

suff..
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Re: M01 #21 [#permalink] New post 21 Aug 2009, 08:43
In the OE of the problem, it is stated that from statement1: We can infer that CAD = CDA, how can this be inferred with the given information.
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Re: M01 #21 [#permalink] New post 21 Aug 2009, 16:25
Economist wrote:
Hi crejoc,

From stmt1:
ACB = 2x, ADC=x

ACD = 180-2x
Now in triangle ACD, ACD + ADC + DAC = 180
=> 180-2x+x+DAC = 180
=> DAC = x = ADC
OR, CAD=CDA


Thanks economist , that didnt strike me.. kudos for your help.
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Re: M01 #21 [#permalink] New post 02 Nov 2010, 07:54
ABC=90;ACB=2CAB => ACB=60;CAB=30 => AB=BC*\sqrt{3} => ABC NOT ISOSCELES.
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Re: M01 #21 [#permalink] New post 07 Nov 2010, 06:57
Since angle ABC=90, to ensure triangle is Isosceles AB = BC (since AC will be hypotenus). therefore angles ACB and CAB must equal 45degrees.

1) tells us nothing about the size of the angle, only a proportion. insuf.
2) ACB = 2CAB, therefore suf (tells us that it can not be isosceles)

therefore B
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Re: M01 #21 [#permalink] New post 07 Nov 2010, 20:31
B

90 + a + 2a = 180 => a =30

scalene triange
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Re: M01 #21 [#permalink] New post 14 Nov 2010, 08:36
B. s1 - if outside angle is 27.5 than inside angle is 45 and hence triangle is isoceles. However for any other value of outside angle this is not possible.
S2 - clearly states triangle cannot be isoceles as the two angles néed to be equal to each other in aright angled triangle

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Re: M01 #21 [#permalink] New post 25 Mar 2012, 07:19
I have a doubt . from crejoc's reply I get that angle CAD= angle ADC . Now considering the triangle ADC the exterior angle ACB = sum of angles( CAD+ CDA) . So angle ACB > angle ADC . Now taking the rule angle opposite to greater side is greater and comparing the angles ACB and ADC we have AB> AC . Hence ABC is not isosceles.

So should n't the answer be D instead of B.... please help...
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Re: M01 #21 [#permalink] New post 26 Mar 2012, 06:27
seofah wrote:
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I'd appreciate some shortcuts to solve this q.

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Tricky one....i did not get the RIGHT answer ....after reading the above explanations, i agree with them that the answer is B
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Re: M01 #21 [#permalink] New post 04 Nov 2013, 05:07
IMO it's B

to prove triangle ABC is isosceles, we need to prove ACB =BCA=45,
This can't be found using option A (on the contrary I came to d conclusion that triangle ACD is isosceles :D)

from option B
it's obvious that in triangle ABC, angle ACB is not equal to BAC
so triangle ABC can't be isosceles
Re: M01 #21   [#permalink] 04 Nov 2013, 05:07
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