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# ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB

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Intern
Joined: 25 Feb 2014
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ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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25 Oct 2014, 03:21
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Question Stats:

69% (01:38) correct 31% (01:36) wrong based on 84 sessions

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ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB. What is the ratio of the area of triangle EFP to that of CDP?

A) 1:3
B) 1:4
C) 1:10
D) 1:9
E) 3:4
[Reveal] Spoiler: OA

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Parallogram.png [ 7.15 KiB | Viewed 1758 times ]

Manager
Joined: 10 Feb 2014
Posts: 116
GMAT 1: 690 Q50 V33
Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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25 Oct 2014, 03:35
Area(EFP)/Area(CFD) = (ratio of sides)^2 = (1/3)^2 = 1/9

Ans. D
Manager
Joined: 18 Aug 2014
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Location: Hong Kong
Schools: Mannheim
Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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26 Oct 2014, 09:48
itzmyzone911 wrote:
Area(EFP)/Area(CFD) = (ratio of sides)^2 = (1/3)^2 = 1/9

Ans. D

Hi,

Kind regards.
Manager
Joined: 10 Feb 2014
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GMAT 1: 690 Q50 V33
Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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26 Oct 2014, 10:24
LaxAvenger wrote:
itzmyzone911 wrote:
Area(EFP)/Area(CFD) = (ratio of sides)^2 = (1/3)^2 = 1/9

Ans. D

Hi,

Kind regards.

This formula takes care of the height.

Ratio of areas of two similar triangles (triangle 1 and triangle 2) = (0.5*base1*height1)/(0.5*base2*height2) = (base1/base2)*(height1/height2)

Remember, in similar triangles ratio of any CORRESPONDING dimensions of the two triangles is constant. This means ratio of corresponding sides = ratio of corresponding heights = ratio of corresponding medians = ratio of corresponding angle bisectors etc. etc. = 1/3

Hence, reqd. ratio = (1/3)*(1/3) = 1/9
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Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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28 Oct 2014, 02:31
Refer diagram below:
Attachment:

Parallogram.png [ 4.6 KiB | Viewed 1490 times ]

Alternate angles are congruent; opposite angles are congruent

DC = 3 times EF

So, height of $$\triangle$$ CDP = 3 times height of $$\triangle$$ EFP

$$\frac{Area \triangle CDP}{Area \triangle EFP} = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$$

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Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB [#permalink]

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04 Feb 2018, 15:15
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Re: ABCD is a parallelogram. E and F are points on AB such that AE=EF=FB   [#permalink] 04 Feb 2018, 15:15
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