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

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Math Expert
Joined: 02 Sep 2009
Posts: 53066
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,  [#permalink]

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09 Oct 2018, 22:24
00:00

Difficulty:

35% (medium)

Question Stats:

77% (02:43) correct 23% (02:41) wrong based on 29 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 ?

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

Attachment:

rectangle.jpg [ 30.48 KiB | Viewed 442 times ]

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

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09 Oct 2018, 23:34
1
Area of parallelogram EBFD = EB x BC
Area of parallelogram AEFC =AE x AD

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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10 Oct 2018, 03:17

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Joined: 09 Mar 2016
Posts: 1284
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,  [#permalink]

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10 Oct 2018, 03:53
Bunuel wrote:

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
In the figure, the areas of parallelograms EBFD and AECF are 3 and 2,   [#permalink] 10 Oct 2018, 03:53
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