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Re: M04, Q7

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Re: M04, Q7 [#permalink] New post 14 Apr 2009, 12:28
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If quadrilateral ABCD is inscribed into a circle, what is the value of \angle BAD ?

1. AC = CD
2. \angle ADC = 70^\circ

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Explanation:

If ABCD is a quadrilateral then:

total of all angle's is 360.

Now, in stmt 1 they are saying that AC (which could be the diameter of the circle) = CD (one of the sides of the quad)
- This helps us in understanding that angle ADC = angle DAC = x
and that x + x + angle ACD = 180
and that angle ACD = 180 - 2x
What we need is angel BAD which is not provided in Stmt 1, thus NS

Stmt 2
gives angel ADC = 70 deg
which does not help us in estimating any other angle since a quadrilateral can be a square, parallelogram or a rhombus so angles can be different

Combining stmt 1 and stmt 2 we get
angle ADC = 70 deg = angle CAD
But we are still missing information about the other angle's i.e.
angle ABC + angle BCD + angle ADC + angle BAD = 360
angle ABC + angle BCD + angle 70 + (angle BAC + angle CAD) = 360
angle ABC + angle BCD + angle 70 + (angle BAC + 70) = 360

Rest of the angles above are still missing. Thus, E
If we were told what kind of a quad it is then it would be easier to find out the angle.
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Re: M04, Q7 [#permalink] New post 22 May 2009, 19:31
Well explained - thank you.

+1

http://www.algebralab.org/lessons/lesso ... istics.xml

I think it is important to remember a quadrilateral can be a shape of four sides. Any shape is fine. A warped Trapezium of obtuse angles would be a qdrltlt as well. Am I right?
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 05:27
dude,ac cannot be diameter of circle because if that were true,adc=90 deg but we cant be sure of it
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 06:31
since ADC = 70 and AC = CD,

DAC = 70 and therefore ACD = 180 - 140 = 40

since inscribed angle of point on circle to A and D is 40,center of circle = O

AOD = 2 * 40 = 80
in triangle AOD, OAD = ODA = 40

BOC = AOD = 80
therefore AOB = DOC = (360 - 80 -80) / 2 = 100

in triangle AOB, OAB = OBA = 50

BAD = OAB + OAD = 40 + 50 = 90

[Reveal] Spoiler:
Answer = C (both statements together)
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 08:09
hey how can we say that centre lies on the dia AC.
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 09:04
how is ac the diameter
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 11:36
The answer is E.

We can just find out all angles of ACD - CAD (70), CDA (70), DCA (40). However, we cannot find any other angle.
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 12:44
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I know the answer.... its E!

Explanation:
I banged my head against the wall after wasting 15 minutes on this question.
I then figured out that I still cannot get to the answer.
So it has to be E.

PS: I think that was wasteful seriously.
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 12:51
adhithya wrote:
how is ac the diameter


my question as well...

before having a kudos/+1 party please answer the eternal question... "how is ac the diameter?"

I had imagined the quadrilateral ABCD inscribed in the the circle at the very top.... let's say in a semi-circle. now how does AC become the diameter???
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Re: M04, Q7 [#permalink] New post 11 Nov 2010, 22:00
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xunil56 wrote:
If quadrilateral ABCD is inscribed into a circle, what is the value of \angle BAD ?

1. AC = CD
2. \angle ADC = 70^\circ

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Explanation:

If ABCD is a quadrilateral then:

total of all angle's is 360.

Now, in stmt 1 they are saying that AC (which could be the diameter of the circle) = CD (one of the sides of the quad)
- This helps us in understanding that angle ADC = angle DAC = x
and that x + x + angle ACD = 180
and that angle ACD = 180 - 2x
What we need is angel BAD which is not provided in Stmt 1, thus NS

Stmt 2
gives angel ADC = 70 deg
which does not help us in estimating any other angle since a quadrilateral can be a square, parallelogram or a rhombus so angles can be different

Combining stmt 1 and stmt 2 we get
angle ADC = 70 deg = angle CAD
But we are still missing information about the other angle's i.e.
angle ABC + angle BCD + angle ADC + angle BAD = 360
angle ABC + angle BCD + angle 70 + (angle BAC + angle CAD) = 360
angle ABC + angle BCD + angle 70 + (angle BAC + 70) = 360

Rest of the angles above are still missing. Thus, E
If we were told what kind of a quad it is then it would be easier to find out the angle.



Hey dude!
you cant assume AC as the diameter of circle.
even if you have assumed this, then the quad will be a Rectangle or Square and its very easy to find the angles with these 2 stmnts.


Sol:
Stmt 2: Angl ADC = 70
stmt 1: AC=DC i.e. angl CAD= angl CDA =70
==> angl DCA =40.
since, Angl ADC + Angl ABC =180;
==> Angl ABC = 110.

we cant find Angl BAD = 70 + Angl BAC, using these 2 condition.
so FInal ANs is E.
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Re: M04, Q7 [#permalink] New post 12 Nov 2010, 00:17
321, how do you get that ABC =110?

We know all angles in the the small triangle ACD. But now B could be ANY point above line AC on the cirlce - does it always end up as the same angle?? If so, please elaborate, how to get that angle.

Regarding the question if AC could or could not be the diameter:
Statement 2 tells you that it is NOT!
Because if AC was the diameter, then ABC and ADC would both be right angles. (but S2 tells you ADC = 70)
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Re: M04, Q7 [#permalink] New post 12 Nov 2010, 01:23
AndreG wrote:
321, how do you get that ABC =110?

We know all angles in the the small triangle ACD. But now B could be ANY point above line AC on the cirlce - does it always end up as the same angle?? If so, please elaborate, how to get that angle.

Regarding the question if AC could or could not be the diameter:
Statement 2 tells you that it is NOT!
Because if AC was the diameter, then ABC and ADC would both be right angles. (but S2 tells you ADC = 70)



Hey,
Theorem : The sum of the opposite angles in cyclic Quadrilateral are supplementary to one another, i.e., their sum is equal to two right angles

so angl ADC + Angl ABC = 180,
n angl ADC = 70.
its the property of cyclic quad.
for more theorem follow the link.
http://www.tutornext.com/cyclic-quadril ... rties/1028
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Re: M04, Q7 [#permalink] New post 16 Nov 2010, 01:26
S1: isosceles triangle, so Angle ADC=DAC. Not sufficient
S2: Angle ADC=70 deg. Not sufficient.
Combining we have Angle ACD=40 (Sum of 3 angles in a triangle ADC=180 Deg).
Now that alternate opp. Angles are equal AC bisects AB & DC, we can say Angle CAB=40.
Therefore Angle BAD=110 deg.

kudos if you like my explanation !
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Re: M04, Q7 [#permalink] New post 15 Nov 2011, 14:42
anandthiru wrote:
S1: isosceles triangle, so Angle ADC=DAC. Not sufficient
S2: Angle ADC=70 deg. Not sufficient.
Combining we have Angle ACD=40 (Sum of 3 angles in a triangle ADC=180 Deg).
Now that alternate opp. Angles are equal AC bisects AB & DC, we can say Angle CAB=40.
Therefore Angle BAD=110 deg.

kudos if you like my explanation !


it is not statex thar it is paralellogram so ur explanation os wrong
E

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Re: M04, Q7 [#permalink] New post 16 Nov 2011, 16:40
I'm glad there are so many opinions on this answer :) I got E. I can only get the measurement for half of angle DAB.
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Re: M04, Q7 [#permalink] New post 03 Jan 2012, 17:46
I drew ABCD as a rectangle with the points on the circle. I ended up with answer D and was surprised to find out it was E.

Since Rectangles have 4 rt angles, for statement 1 I drew a diagonal line through the rectangle to form 2 rt triangles, and since ac = cd, I thought it was isosceles thus I had <ADC and < DAC = 45 each therefore able to answer the original question.

For statement 2, Since <ADC=70, I was able to use a similar method and turned the rectangle into 2 rt triangles. I added 90 and 70 and then subtracted from 180 to find <DAC. I then was able to find out the value for <BAD by subtracting <DAC from 90, thus I labeled statement 2 as sufficient.
I was wondering what I did wrong since I was off by a lot.
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Re: M04, Q7 [#permalink] New post 04 Jan 2012, 16:14
A quick question:

For which quadrilaterals do diagonals bisect the vertex angles?
Re: M04, Q7   [#permalink] 04 Jan 2012, 16:14
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