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In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,

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Math Expert
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In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,  [#permalink]

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New post 02 Apr 2019, 23:25
1
1
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

88% (01:57) correct 13% (02:29) wrong based on 32 sessions

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Re: In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,  [#permalink]

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New post 03 Apr 2019, 12:03
1
AD = BC = CE = x = 5
AB = CD = y = 10
Then, BE = \(5\sqrt{2}\) (45-45-90 triangle) = AF
Now, FD = \(\sqrt{50-25}\) = 5
Area of AFD = 1/2*5*5 = 12.5

C is the answer.
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Re: In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,  [#permalink]

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New post 06 Apr 2019, 13:27
Bunuel wrote:
Image
In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5, and y = 10, then what is the area of ∆AFD ?

(A) 2.5

(B) 5

(C) 12.5

(D) 50

(E) 50 + 5y

Attachment:
#GREpracticequestion In the figure, ABCD.jpg


Alternate Soln.
\(\Delta ADF \,\, and \,\,\Delta BCE\, are\,\, similar \\
AD=BC=5\\
DF=CE=5\\
Area \,\,of \,\Delta ADF \Rightarrow \frac{1}{2}*5*5 =12.5\)
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Re: In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,  [#permalink]

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New post 06 Apr 2019, 21:37
tri.ADF and tri.BCE are congruent by AAA property.
Hence, sides will be equal in length.

Area = 25/2= 12.5

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Re: In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,   [#permalink] 06 Apr 2019, 21:37
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In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,

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