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

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

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New post 09 Oct 2018, 23:24
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (02:33) correct 19% (02:27) wrong based on 38 sessions

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Joined: 04 Jun 2015
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Re: In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,  [#permalink]

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New post 10 Oct 2018, 00:34
1
Area of parallelogram EBFD = EB x BC
Area of parallelogram AEFC =AE x AD
BC=AD = y

Adding both of the area
EB * y + AE * y =3+2
(EB + AE) *y =5 = AB *y
So area of reactangle ABCD =5
Option :C

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

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New post 10 Oct 2018, 04:17
chetan2u your approach please

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

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New post 10 Oct 2018, 04:53
Bunuel wrote:
Image
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2, respectively. What is the area of rectangle ABCD ?

(A) 3
(B) 4
(C) 5
(D) \(4 \sqrt{3}\)
(E) 7

Attachment:
rectangle.jpg


Area of Parallelogram - base * height


hence if area of EBFD is 3, that means that base is 3 and height is 1 ---> 3*1=3

simiralry area of AECF is AECF, meaning base is 2 and height is 1 ---- > 2*1 =2

Area of rectangale base is 2+3 =5 and height is 1, so 5*1 = 5

so answer is C i think :)
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In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,   [#permalink] 10 Oct 2018, 04:53
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In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,

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