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Geometry 7

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Geometry 7 [#permalink] New post 02 Oct 2009, 09:57
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Diagram not to scale
Q.
ABCD is a trapezium in which AB is parallel to DC, AD = BC, AB= 6 cm, AB=EF and DF=EC. If the two lines AF and BE are drawn so that area of ABEF is half of ABCD.
Find DF/CD.
1. 1/4
2. 1/3
3. 2/5
4.1/6
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Manager
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Re: Geometry 7 [#permalink] New post 13 Oct 2009, 16:00
virtualanimosity wrote:
Diagram not to scale
Q.
ABCD is a trapezium in which AB is parallel to DC, AD = BC, AB= 6 cm, AB=EF and DF=EC. If the two lines AF and BE are drawn so that area of ABEF is half of ABCD.
Find DF/CD.
1. 1/4
2. 1/3
3. 2/5
4.1/6


1/3
as per diagram and info AB=EF=6 cm and DF = EC = x(say)
then area of ABEF is 6*h (hieght) = 1/2 [1/2*( 6+6+2x)*h]
we get x= 6 so DF/ CD = 6/18 =1/3
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Re: Geometry 7 [#permalink] New post 13 Oct 2009, 17:31
My Ans. is 2

1/2 (6+CD)h = 2* 1/2 (6+6)h

solving for CD, we get 18, since EF=6 and DF = EC , DF, EC must be 6 each.

hence DF/CD=6/18=1/3

matrix is always ahead of me and almost always right :) :)
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Re: Geometry 7 [#permalink] New post 13 Oct 2009, 22:42
srini123 wrote:
My Ans. is 2

1/2 (6+CD)h = 2* 1/2 (6+6)h

solving for CD, we get 18, since EF=6 and DF = EC , DF, EC must be 6 each.

hence DF/CD=6/18=1/3

matrix is always ahead of me and almost always right :) :)

:lol:
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Re: Geometry 7 [#permalink] New post 14 Oct 2009, 02:35
AB=EF=6
DF=EC=x cm

Area of ABCD = 2 ( Area of ABEF )
1/2*(6+6+x+x) = 2 ( 1/2*12*h )
x=6
thus DF/CD = 6/18 = 1/3
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Re: Geometry 7   [#permalink] 14 Oct 2009, 02:35
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