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# Math Question

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Senior Manager
Joined: 05 Oct 2008
Posts: 263

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17 Oct 2008, 08:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can anyone help to solve the 2 questions in the file. I don't know how to upload an image and hence put it in a doc.

One question is in one of the replies below and the other is in the document.

Figure in document atached

In the figure above, ABE = CBE = 30°. If CD = 10 and DE = 6, what is the perimeter of quadrilateral ABCD?
40
48
52
54
72

Last edited by study on 19 Oct 2008, 09:19, edited 5 times in total.
VP
Joined: 30 Jun 2008
Posts: 1022

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17 Oct 2008, 08:07
study wrote:
Can anyone help to solve the questions in the file. I don't know how to upload an image and hence put it in a doc.

Attachment:
Math QS.doc

U can upload an image just the way you upload a doc. If you upload a pic, you will generally receive more responses ....
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Senior Manager
Joined: 05 Oct 2008
Posts: 263

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17 Oct 2008, 08:09
dont have it in the form of a picture...
VP
Joined: 30 Jun 2008
Posts: 1022

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17 Oct 2008, 08:11
Is angle DFE = 60° ?
(1) Line segment EG line bisects line segment DF
(2) Angle EDF = 60°

Is it C ?
Attachments

q1.jpg [ 3.56 KiB | Viewed 1427 times ]

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"You have to find it. No one else can find it for you." - Bjorn Borg

SVP
Joined: 17 Jun 2008
Posts: 1507

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17 Oct 2008, 13:23
E for the first one.

Stmt1 does not give any clue about angle DFE. Insufficient.
Stmt2 gives angle EDF.But, there is no clue about other angles. Also, EG is just a bi-sector and not a perpendicular bisector of DF.

For the second one, 52.
In triangle CDE, EC will be 8. Now, BEC is a 30-60-90 triangle, hence BC = 16. similarly, AB = 16 and AD = 10.

Hence, perimeter = 16 + 16 + 10 + 10 = 52.
VP
Joined: 17 Jun 2008
Posts: 1329

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17 Oct 2008, 13:47
scthakur wrote:
E for the first one.

Stmt1 does not give any clue about angle DFE. Insufficient.
Stmt2 gives angle EDF.But, there is no clue about other angles. Also, EG is just a bi-sector and not a perpendicular bisector of DF.

For the second one, 52.
In triangle CDE, EC will be 8. Now, BEC is a 30-60-90 triangle, hence BC = 16. similarly, AB = 16 and AD = 10.

Hence, perimeter = 16 + 16 + 10 + 10 = 52.

ABC is a equilateral triangle with all 60 deg angles !!!
hence CE=8 (calculated from DE and CD of right trangle)=> AC=16=AB=BC

AE=CE=8 (alitude of equilateral triangle bisects the base)
now CD=10 ad AD=10 (from AE and ED of right trianngle AED)
HENCE 10+10+16+16=52 = perimeter
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Joined: 17 Jun 2008
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17 Oct 2008, 13:54
spriya wrote:
ABC is a equilateral triangle with all 60 deg angles !!!
hence CE=8 (calculated from DE and CD of right trangle)=> AC=16=AB=BC

AE=CE=8 (alitude of equilateral triangle bisects the base)
now CD=10 ad AD=10 (from AE and ED of right trianngle AED)
HENCE 10+10+16+16=52 = perimeter

How is ABC an equilateral triangle? Question does not mention this.
Director
Joined: 23 May 2008
Posts: 762

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17 Oct 2008, 15:49
scthakur wrote:
spriya wrote:
ABC is a equilateral triangle with all 60 deg angles !!!
hence CE=8 (calculated from DE and CD of right trangle)=> AC=16=AB=BC

AE=CE=8 (alitude of equilateral triangle bisects the base)
now CD=10 ad AD=10 (from AE and ED of right trianngle AED)
HENCE 10+10+16+16=52 = perimeter

How is ABC an equilateral triangle? Question does not mention this.

ABC is equilateral because angle ABE= angle CBE = 30 degrees...we have two 30 , 60 , 90 triangles on top
Senior Manager
Joined: 05 Oct 2008
Posts: 263

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17 Oct 2008, 22:42
Thanks for the explanation. Actually a 30/60/90 triangle has standard triangles such as
5, 12, 13
3, 4, 5
8, 15, 17
so how did you get 16 instead of 17?

Thanks.
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown

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18 Oct 2008, 12:11
Question 1 : E

@study
Question 2 :
30:60:90 triangle will give you 1:sqrt(3) : 2

So, taking everyone's explanation, EC:BE:BC :: 8:sqrt(3)*8:16

That is why you get 16
Senior Manager
Joined: 05 Oct 2008
Posts: 263

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18 Oct 2008, 21:37
I know how you got 16. But my question is when do you know which rule to use. For eg. If you have a 30, 60, 90 triangle with side 5. Would you take the hypotenuse to be 10 or 13 (5,12,13 triangle)?
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown

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19 Oct 2008, 01:30
oh ok, the way it works is

30:60:90
1:sqrt(3):2

So, the side opposite 30 will be 1
side opposite 60 will be sqrt(3)
side opposite 90 will be 2

It is the same with 45:45:90::1:1:sqrt(2)

Re: Math Question   [#permalink] 19 Oct 2008, 01:30
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# Math Question

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