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# In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r

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Director
Joined: 23 Feb 2015
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GMAT 1: 720 Q49 V40
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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29 Sep 2016, 01:42
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78% (02:32) correct 22% (01:59) wrong based on 70 sessions

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In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 respectively. What is the area of rectangle ABCD?
Thanks Expert...

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Director
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In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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28 Feb 2017, 04:10
3
2
Area of EBFD = Base *height = FD *BC = 3
Area of AECF = Base *Height = FC*BC = 2
Area of ABCD = CD* BC
= (CF + FD)*BC
= CF*BC + FD*BC
= Area of AECF + Area of EBFD
= 2 +3 =5. Option C
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Director
Joined: 23 Feb 2015
Posts: 842
GMAT 1: 720 Q49 V40
Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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29 Sep 2016, 01:55
Hi Expert,
The area of AECF is FC*AD. Also, the area of EBFD is DF*AD. Now, Summing the two areas yields:
------> FC*AD+DF*AD
------> AD(FC+DF)
------> DC*AD, which is the area of ABCD.
Here, my question is:
Here is given the areas of AECF and EBFD. The two areas did not include the red part. But, when we calculate this two parallelograms areas, it automatically calculate the red part of the areas. if EBFD is 3 and AECF is 2, then their whole areas will be 2+3=5, which did not include red part, right? But, when we ADD red part, its (ABCD) area is 5. My question is: Des red part not have any value?
Thanks...
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Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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20 Aug 2017, 01:56
Take triangle EBC and ADF. these are equilateral traingles. Since EB=DF and BC =height of parallelogram = EB

Area of EBFD=3 = EB x h= EB x BC
Area of Triangle EBC= 1/2 x EB x BC= 3/2
Area of triangle ADF= Area of Triangle EBC= 3/2
Total area of rectangle = Area of EBC +Area of ADF + Area of parellogram AEFC
=3/2 +3/2 + 2= 3+2= 5

Hope it helps.
Kudos if it does,please.
Director
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Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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20 Aug 2017, 03:39
Area of llgm = base * ht
EBFD = 3 = BC * BE---1
AECF = 2 = BC * AE---2

Add 1 and 2

BC * AB = 5
Area of rectangle ..

Red part is distraction .... :P
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Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r  [#permalink]

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20 Aug 2017, 07:58
iMyself wrote:
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 respectively. What is the area of rectangle ABCD?
Thanks Expert...

This question looks difficult but it is very easy
let the height of the parallelogram be h no notice that we have to find the area of the rectangle .
That means both the given parallelogram are between parallel lines and their height will be same
So area of parallelogram EBFD =3 =a*h.......1
area of parallelogram AECF=2=b*h........2
Now notice the equations 1 and 2
We have h=1
So area of rectangle =(a+b)*h=(3+2)*1=5
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Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r   [#permalink] 20 Aug 2017, 07:58
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# In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 r

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