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# In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d

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Director
Joined: 07 Jun 2004
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Location: PA
In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 10:49
2
5
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Difficulty:

85% (hard)

Question Stats:

54% (08:12) correct 46% (02:32) wrong based on 78 sessions

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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?

(1) Angle DAB = 45 degrees

(2) Angle ADC + angle BCD = 205 degrees

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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 11:44
I got C

Thanks
Director
Joined: 07 Jun 2004
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 13:51
The OA is D without explanation i think its a TYPO

can you let us know how you got C
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Math Expert
Joined: 02 Sep 2009
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 14:21
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rxs0005 wrote:
The OA is D without explanation i think its a TYPO

can you let us know how you got C

No typo OA is correct.

Important property of polygons: Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides.

So for quadrilateral sum of the angles equals to $$180(4-2)=360$$ degrees.

Given: $$\angle{ABC}=110$$ and $$\angle{AED}=100$$. Question: $$\angle{BCD}-\angle{ADE}=?$$.

Now as $$\angle{ABC}=110$$ and $$\angle{AED}=100$$ and the sum of all angles is 360 degrees then the sum of other two angles in quadrilateral must be $$\angle{BCD}+\angle{BAE}=360-(110+100)=150$$.

(1) $$\angle{DAB}=45$$ --> $$\angle{BAE}=\angle{DAB}+\angle{DAE}=45+\angle{DAE}$$, but $$\angle{DAE}=180-100-\angle{ADE}=80-\angle{ADE}$$ (the sum of interior angles in triangle ADE equals to 180 degrees) --> $$\angle{BAE}=45+80-\angle{ADE}=125-\angle{ADE}$$. So $$\angle{BCD}+\angle{BAE}=150$$ becomes $$\angle{BCD}+125-\angle{ADE}=150$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.

(2) $$\angle{BCD}+\angle{ADC}=205$$ --> $$\angle{ADC}=180-\angle{ADE}$$ (straight line) --> $$\angle{BCD}+180-\angle{ADE}=205$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.

Hope it's clear.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 16:35
good Qn.
Thanks Bunuel.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 16:40
thanks i now get it

i was taking interior angles as 180 which is not necessarily true
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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10 May 2018, 09:29
Hii
Please observe that both statement 1 & 2 gives same results.

(1) Angle DAB = 45 degrees
if in quadrilateral, Angle DAB = 45 (St1), Angle ABC= 110 (Question stem)
then as the sum of Angles in a Quadrilateral is 360,
Angle ADC + angle BCD = 205 degrees= Statement 2.
(you need not calculate this, you can infer that Angle ADC + angle BCD = can be calculated from St 1)

(2) Angle ADC + angle BCD = 205 degrees

Since, both statement 1 & 2 gives same results Answer is D or E.
Eliminate all other choices.

We need to check any one of the statements, Let check using statement1

Now , we can solve this question using construction approach:
See the Sketch.
Attachment:

WhatsApp Image 2018-05-10 at 21.56.11.jpeg [ 67.38 KiB | Viewed 289 times ]

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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d   [#permalink] 10 May 2018, 09:29
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