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In the figure, the areas of parallelograms EBFD and AECF are
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01 Oct 2018, 13:44
01 Oct 2018, 13:44
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89% (01:53) correct
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#GREpracticequestin In the figure, the areas of parallelograms EBFD .jpg [ 30.48 KiB | Viewed 4481 times ]
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
Re: In the figure, the areas of parallelograms EBFD and AECF are
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02 Oct 2018, 06:12
02 Oct 2018, 06:12
solution pls??
Re: In the figure, the areas of parallelograms EBFD and AECF are
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03 Oct 2018, 13:31
03 Oct 2018, 13:31
3
Expert Reply
Let DF be x and FC be y and the height of the rectangle (BC) be h. Asshown in figure below:
Inkedrectangle_LI.jpg [ 567 KiB | Viewed 5660 times ]
Now area AECF =2 and writing it in terms of our variable we get
area AECF =2= \(y \times h\).
Similarly area EBFD =3 and writing it in terms of our variable we get
area EBFD =3= \(x \times h\).
Now area of the rectangle is clearly =\((x+y) \times h\).
Clearly if we add the area of EBFD and AECF we get the area of rectangle = \((x+y) \times h= x \times h + y \times h = 3+2= 5\)
Hence option C
Attachment:
Inkedrectangle_LI.jpg [ 567 KiB | Viewed 5660 times ]
Now area AECF =2 and writing it in terms of our variable we get
area AECF =2= \(y \times h\).
Similarly area EBFD =3 and writing it in terms of our variable we get
area EBFD =3= \(x \times h\).
Now area of the rectangle is clearly =\((x+y) \times h\).
Clearly if we add the area of EBFD and AECF we get the area of rectangle = \((x+y) \times h= x \times h + y \times h = 3+2= 5\)
Hence option C
