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# Geometry II = #12

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Geometry II = #12 [#permalink]  05 Aug 2009, 17:41

In quadrilateral $$ABCD$$ , $$AB = CD$$ and $$BC = AD$$ . If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$ ?

(C) 2008 GMAT Club - Geometry - II#12

* 30 degrees
* 50 degrees
* 70 degrees
* 100 degrees
* 120 degrees

Given the question, doesn't that violate the rules of a parallelogram? I just can't seem to draw this figure out as a parallelogram. somebody help please?
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Re: Geometry II = #12 [#permalink]  06 Aug 2009, 10:55
5
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bipolarbear wrote:
I thought that a parallelogram had to have 2 adjacent angles add up to 180. Can someone attach a drawing of this figure please?

OA is D.

Hey! I got D as an answer as well and I think I can explain it. I attached a drawing I created in Paint so hopefully that'll help too.

Here's how I broke it down:
-They gave two angles, the 30 (CBD) and 80 (BAD).
-I drew in the CBD angle to make it easier to visualize.
-In a parallelogram, when one line (BD) connects the two parallel lines (AD and BC), it creates congruent alternate interior angles (the 30s and the X's). Angles diagonal from eachother also congruent, which is how I got the other 80 (BCD). You can also see that CBA and ADC (the one in question) are also congruent since both of them equal X+30. So we just need to figure out what X is so we can add it to 30 to get the whole angle.
-All the angles inside a parallelogram are supposed to have a sum of 360. So if you take 360 and subtract all the known values (80, 80, 30, 30) you should get a difference of 140.
-Therefore the two remaing angles (the X's) should equal 140. If you set it up as 2x=140 and then divide it out, you'll see that X=70.
-Then X+30 becomes 70+30 which equals 100.

I hope my explanation helps.... and is right Yay for 1st post!
Attachment:

GeoII-12.JPG [ 11.59 KiB | Viewed 1690 times ]
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Re: Geometry II = #12 [#permalink]  19 Feb 2013, 04:23
2
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Expert's post
Sachin9 wrote:
Bunuel,
In a parallelogram, do we have sum of all interior angles = 360?

do we have any other propery apart from opposite sides and opposite angles being equal?

The sum of interior angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles). So, the sum of the interior angles of any quadrilateral, including parallelogram, is 180*2=360.

Parallelogram A quadrilateral with two pairs of parallel sides.

Properties and Tips
• Opposite sides of a parallelogram are equal in length.
• Opposite angles of a parallelogram are equal in measure.

• Opposite sides of a parallelogram will never intersect.
• The diagonals of a parallelogram bisect each other.
• Consecutive angles are supplementary, add to 180°.
The area, $$A$$, of a parallelogram is $$A = bh$$, where $$b$$ is the base of the parallelogram and $$h$$ is its height.
• The area of a parallelogram is twice the area of a triangle created by one of its diagonals.

A parallelogram is a quadrilateral with opposite sides parallel and congruent. It is the "parent" of some other quadrilaterals, which are obtained by adding restrictions of various kinds:
• A rectangle is a parallelogram but with all angles fixed at 90°
• A rhombus is a parallelogram but with all sides equal in length
• A square is a parallelogram but with all sides equal in length and all angles fixed at 90°

For other properties check polygons chapter of Math Book: math-polygons-87336.html
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Re: Geometry II = #12 [#permalink]  06 Aug 2009, 09:37
1
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I thought that a parallelogram had to have 2 adjacent angles add up to 180. Can someone attach a drawing of this figure please?

OA is D.
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Re: Geometry II = #12 [#permalink]  05 Aug 2009, 19:32
bipolarbear wrote:

In quadrilateral $$ABCD$$ , $$AB = CD$$ and $$BC = AD$$ . If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$ ?

(C) 2008 GMAT Club - Geometry - II#12

* 30 degrees
* 50 degrees
* 70 degrees
* 100 degrees
* 120 degrees

Given the question, doesn't that violate the rules of a parallelogram? I just can't seem to draw this figure out as a parallelogram. somebody help please?

not sure how this one works too...
.
but i wanted to know if you were to get any of these questions of a quadrilateral ABCD = how would u determine which points are A, B, C, D? Is it supposed to decided in a clockwise fashion?
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Re: Geometry II = #12 [#permalink]  05 Aug 2009, 21:47
bipolarbear wrote:

In quadrilateral $$ABCD$$ , $$AB = CD$$ and $$BC = AD$$ . If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$ ?

(C) 2008 GMAT Club - Geometry - II#12

* 30 degrees
* 50 degrees
* 70 degrees
* 100 degrees
* 120 degrees

Given the question, doesn't that violate the rules of a parallelogram? I just can't seem to draw this figure out as a parallelogram. somebody help please?

Parallelogram = a quadilateral with opposit equal and parallel sides.
How the rules of a parallelogram are violated?

Did you get D. 100 degrees?
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Re: Geometry II = #12 [#permalink]  05 Aug 2009, 22:05
i got the answer as D>100 degrees, by drawing a figure, and simply using the properties of parallel lines...

wats the OA..
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Re: Geometry II = #12 [#permalink]  06 Aug 2009, 11:23
Expert's post
TheCojones wrote:

al 140. If you set it up as 2x=140 and then divide it out, you'll see that X=70.
-Then X+30 becomes 70+30 which equals 100.

I hope my explanation helps.... and is right Yay for 1st post!

Kudos! and welcome to GMAT Club!
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Re: Geometry II = #12 [#permalink]  06 Aug 2009, 11:43
TheCojones wrote:
Hey! I got D as an answer as well and I think I can explain it. I attached a drawing I created in Paint so hopefully that'll help too.

Wow, thanks! Now it all makes sense . kudos to you! Hopefully you can stick around here and make use of these awesome forums (ehhh, not just for my sake ) !
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Re: Geometry II = #12 [#permalink]  06 Aug 2009, 11:59
hehe thanks guys! I just found this site today and have already been stuck on it for a good couple of hours. I'm sure I'll be around
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Re: Geometry II = #12 [#permalink]  19 Feb 2013, 03:28
Bunuel,
In a parallelogram, do we have sum of all interior angles = 360?

do we have any other propery apart from opposite sides and opposite angles being equal?
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Re: Geometry II = #12 [#permalink]  06 Feb 2014, 11:50
Bunuel wrote:
Sachin9 wrote:
Bunuel,
In a parallelogram, do we have sum of all interior angles = 360?

do we have any other propery apart from opposite sides and opposite angles being equal?

The sum of interior angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles). So, the sum of the interior angles of any quadrilateral, including parallelogram, is 180*2=360.

Parallelogram A quadrilateral with two pairs of parallel sides.

Properties and Tips
• Opposite sides of a parallelogram are equal in length.
• Opposite angles of a parallelogram are equal in measure.

• Opposite sides of a parallelogram will never intersect.
• The diagonals of a parallelogram bisect each other.
• Consecutive angles are supplementary, add to 180°.
The area, $$A$$, of a parallelogram is $$A = bh$$, where $$b$$ is the base of the parallelogram and $$h$$ is its height.
• The area of a parallelogram is twice the area of a triangle created by one of its diagonals.

A parallelogram is a quadrilateral with opposite sides parallel and congruent. It is the "parent" of some other quadrilaterals, which are obtained by adding restrictions of various kinds:
• A rectangle is a parallelogram but with all angles fixed at 90°
• A rhombus is a parallelogram but with all sides equal in length
• A square is a parallelogram but with all sides equal in length and all angles fixed at 90°

For other properties check polygons chapter of Math Book: math-polygons-87336.html

In this questions was it sufficient on the basis of the fact that two opposites sides of a quadliteral are equal so it must be a Parallelogram?
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Re: Geometry II = #12   [#permalink] 06 Feb 2014, 11:50
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# Geometry II = #12

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