Geometry area - please help

Question

Hi all,
I am confused in a problem. Please help me.

In rectangle ABCD, side BC is twice as long as side AB. Point E is on side CD with DE:EC = 1:2. Segments BE and AC intersect at point F. The area of quadrilateral AFED is what fraction of the area of rectangle ABCD? Express your answer as a common fraction.

ans : 1/6. Please tell me how to do it.

Thanks,
Geet

resh · Answer

I just tried plotting it, not getting it either.

resh · Answer

The sides are -

AB=CD=x
BC=AD=2x
DE=x/3;EC=2x/3

Not sure, how to proceed from here.

geet1234 · Answer

i did the same thing and me too not getting..

anyone else please help

aim730 · Answer

Geet it is not an Official GMAT question What is the source of this ques.

amp0201 · Answer

what is the source of question?

prashantbacchewar · Answer

geet1234 can you please provide detailed answer for this question. Does not seem to be GMAT question but still will be helpful to build the concepts.

utin · Answer

not able to get anywhere...

gmat1011 · Answer

i get 2/3 if i assume a line joining F to a point on EC (say G) when drawn parallel to BC results in DE = EG = GC (but this step and am not being able to base on any property)

I am not sure what property of rectangles/diagonals is being tested here, if this is a genuine Q, but I think it comes down to the ratio in which AC and BE divide each other. Diagonals of a rectangle of course bisect each other and are equal...

shrouded1 · Answer

Very good question, but I doubt if (a) this is a GMAT level question, its much harder and (b) if 1/6 is the answer. See below my detailed answer

This assumption is incorrect

The solution

Let AB=CD=x
Then BC=AD=2x

In triangle FCE construct the perpendicular from vertex F onto side EC, let it intersect the side EC at G.
Lets call angle FCG  	heta_1 
Lets call angle FEG  	heta_2

FG = GC *  Tan(	heta_1)  = GE *  Tan(	heta_2)

We also know that  tan(	heta_1) = \frac{AD}{DC} = 2 
and that  tan(	heta_2) = \frac{BC}{EC} = \frac{2x}{2x/3} = 3

Lets say GC =  \alpha  and GE =  \beta

Therefore  \alpha * tan(	heta_1) = \beta * tan(	heta_2) 
OR  \alpha * 2 = \beta * 3

Also GE + GC = EC = 2/3 * CD = 2/3 * x

Therefore  \alpha + \beta = 2/3 * x

Solving these two equations in  \alpha  and  \beta , we get  \beta = 4/15 x ; \alpha = 6/15 x

Area of AFED = Area (ACD) - Area (FCE) = 0.5 * AD * CD - 0.5 * EC * FG = x^2 - 0.5 * 2/3 x * EG * Tan(	heta_2) = x^2 - 0.5 * 2/3x * 4/15x * 3 = x^2 (1 - \frac{4}{15}) = x^2 * \frac{11}{15}

Area of ABCD = x * 2x =  2x^2

Therefore,

\frac{Area(AFED)}{Area(ABCD)} = \frac{11}{30}

saxenashobhit · Answer

AB = 3x
BC = 6x
CE = x
ED = 2x

Calculating
AC = \(3\sqrt{5}\)x
BE = \(\sqrt{37}\)x

AFB is similar to triangle CFE and ratio of sides is 1:3
So CF = \(3\sqrt{5}x/4\)
and EF = \(\sqrt{37}x/4\)
Semiperimeter = ..
Area of triangle CEF is \sqrt{1}

Attachment:

File comment: Rectangle diagram

Rectangle.jpg [ 8.68 KiB | Viewed 2978 times ]

And I lost my patience

hemanthp · Answer

What shrouded did was pretty good! I will extend on it and try to make it simpler. In the attached figure (excuse my artistic talent): Rectangle.jpg Assume that the side BC = 6 units and AB = 3 units. This makes DE=1 and EC=2.units. So Draw a perpendicular onto CD from F. Call it FG. So in Tan(x) = CG/FG = CD/AD (parallel lines and X will be the corresponding angle) = 3/6=1/2. So that gives us CG = 1/2*FG -> (1) And then Tan(y) = EG/FG = EC/BC = 2/6 = 1/3 That gives us EG = 1/3*FG ->(2) We know that EG+GC = EC = 2 From (1) and (2): So 2 = (1/2+1/3)FG = 5/6*FG. So FG = 12/5 So now the area of triangle EFC = 1/2*12/5*2 = 12/5 Area of triangle AED = 1/2*6*3 = 9. So quad area = 9 - 12/5 = 33/5 So ratio = (33/5)/18 = 33/90 = 11/30. --- I deserve Kudos ..at least for my art

Geometry area - please help

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