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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =

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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

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New post Updated on: 09 Jul 2013, 06:50
1
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11
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A
B
C
D
E

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  95% (hard)

Question Stats:

44% (02:01) correct 56% (02:44) wrong based on 160 sessions

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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.
Attachment:
1.JPG
1.JPG [ 5.35 KiB | Viewed 10095 times ]


A. 1/3
B. 13/16
C. 5/9
D. 1/2

Originally posted by RaviChandra on 14 Sep 2010, 23:58.
Last edited by Bunuel on 09 Jul 2013, 06:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: geomertry Q3&4  [#permalink]

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New post 14 Sep 2010, 23:59
1
In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16

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2.PNG
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Re: geomertry Q3&4  [#permalink]

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New post 15 Sep 2010, 00:54
RaviChandra wrote:
ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.
Attachment:
The attachment 1.JPG is no longer available


a. 1/3
b. 13/16
c. 5/9
d. 1/2


Rotate the triangle so that CA to be the base. Now, triangle CDA has the same base as CBA and half of its height ("read" the diagram for explanation), which means that area of CDA will be half of the area of CBA, so 1/2.
Attachment:
7.PNG
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Answer: D.

RaviChandra these are not GMAT questions, so I wouldn't worry about them too much and definitely wouldn't spend much time on them.
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Re: geomertry Q3&4  [#permalink]

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New post 15 Sep 2010, 01:39
Wht is answer of the 2nd question ???

24 ?
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Re: geomertry Q3&4  [#permalink]

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New post 15 Sep 2010, 02:03
2
RaviChandra wrote:
In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16

Attachment:
2.PNG


Note that AED is similar to ABC with each side as 1/3rd of the original
=> Area(AED) = 1/9 * Area(ABC) =24

Area(ABD) = Area(ACE) = 1/3 * Area(ABC) = 72
[same height, 1/3rd base]

Area(ABD) = Area(ACE)
=> Area(ABD) - Area(AEDF) = Area(ACE) - Area(AEDF)
=> Area(BEF) = Area(CDF) = x (say)

Finally note that DEF is similar to BFC with one side in the ratio of 1:3
Thus area(DEF) = 1/9 * area(BFC) = y (say)

Area(EDC) = Area(EDF)+Area(DFC) = x + y
ALSO
Area(EDC) = Area(AEC) - Area(AED) =72 - 24 = 48
So x+y = 48

Area(BEDC) = Area(BEF) + Area(CDF) + Area(EDF) + Area(BFC) = x+x+y+9y
ALSO
Area(BEDC) = Area(ABC)-Area(AED)=216-24=192
So x+5y=96

Subtracting the 2 equations
4y=48
y=12

Hence, Area(DEF)=12

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Re: geomertry Q3&4  [#permalink]

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New post 15 Sep 2010, 05:02
Wow... thks a ton.

I had guessed it to be 24 - Thinking T. EFD should be of area 1/9th of main traingle.

If I knew T. AED = 1/9 T. ABC, 12 would have been my guess, as it (T. EFD) looks half the size of T. AED
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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

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New post 28 Sep 2015, 06:38
1
Whoa!!! A tough one

A small trick that helped me to solve it in 1:37

Imagine that the triangle ABC is an isosceles triangle with AB = BC. (Reason: All the mentioned conditions are satisfied so area ratio should also hold for Isosceles triangle)

Redraw it in your notepad, you will find symmetry. D is just in middle of AF and DE. You will figure out that D is a mid-point so its height should be half. Now if height of triangle is reduced to half, area of triangle will also reduce to half of its original value (Area = 0.5* base * height).

Hence D.
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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

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