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

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Joined: 02 Sep 2009
Posts: 54371
In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5,  [#permalink]

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02 Apr 2019, 23:25
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Difficulty:

15% (low)

Question Stats:

87% (01:39) correct 13% (02:43) wrong based on 23 sessions

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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 [ 20.56 KiB | Viewed 350 times ]

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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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03 Apr 2019, 12:03
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

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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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06 Apr 2019, 13:27
Bunuel wrote:

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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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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