In the figure above, the quadrilateral ABCD is a rectangle and AE= EB.

Question

GMATBusters’ Quant Quiz 2.0 Question -8

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In the figure above, the quadrilateral ABCD is a rectangle and AE= EB. What is the ratio of the area of shaded triangle DFC and the area of the rectangle ABCD?
A. \(\frac{1}{2}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{5}\)
E. Cannot be determined

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Official Answer

B

CareerGeek · Answer

Let the length and breadth be ‘l’ and ‘b’ respectively

We know that triangles AFE & DFC are similar and
AE = 1/2(DC)
—> All the corresponding sides and the even perpendicularly drawn from F to AE & DC will be in the same ratio (similar triangle property)
Let the perpendicular from F to AE = x
—> perpendicular from F to DC = 2x
—> Also, x + 2x = b
—> x = b/3

So, Area of triangle DFC = 1/2*2x*DC = x*l = lb/3
Area of rectangle = lb
—> Required ratio = (lb/3)/lb = 1/3

IMO B

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numb007 · Answer

Ans:B
AFE and DFC are similar triangle because of all the same angles.
Let AB=l
Let AD=b
AD is divided in x and b-x at the point projected at right angle on AD from point F.
Hence Area of AFE/Area of DFC= (AE/DC)^2
hence (x/2)/b-x=1/4
x=b/3
HEnce Area of DFC/Area of ABCD=(1/2*(b-x)*l))/(l*b)=1/3

rvarora · Answer

Answer is 1/3 that is B
Attaching the solution herewith

willacethis · Answer

Since the given figure is a rectangle, we can change the ratio without affecting the required condition as mentioned in the question by stretching the rectangle vertically or horizontally (as shown in 2 diagrams).

Hence, the answer cannot be determined = option E

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ANSWER 2.png [ 443.02 KiB | Viewed 8832 times ]

Hence, the answer cannot be determined = option E

pk123 · Answer

In the figure above, the quadrilateral ABCD is a rectangle and AE= EB. What is the ratio of the area of shaded triangle DFC and the area of the rectangle ABCD?

Sine E is the midpoint , AE=EB=AB/2

Lets say AB= L (length of rectangle)
BC = AD = B (Width of rectangle)

Lets draw a line parallel to AD and BC passing through F meeting CD at G and AB at H
say FG=w and FH=x
 and w+x= B = AD=BC....(width of rectangle)

Area of triangle (DFC)= 1/2*w*CD =1/2*w*L = wL/2
area DFC/2 = wL/4 --------(1)
Area of triangle (AEF) = 1/2*x*AE = 1/2*x*L/2 = xL/4 ---------(2)

Combining both (1) & (2)
area DFC/2 + area AEF = (W+x)*L/4 = LB/4 
so , Area AEF = LB/4 - Area DFC/2 -------(3)

Area ADC = 1/2*DC*AD = 1/2*L*B = LB/2 = Area DFC+Area ADF
Area ADF = LB/2 - Area DFC ---------(4)
 
Area of ADF + Area AEF = Area ADE = 1/2*B*L/2= LB/4 
LB/2 - Area DFC + LB/4 - Area DFC/2 = LB/4
hence 3*Area DFC/2 = LB/2
Area DFC= LB/3

So Area DFC/ Area ABCD = LB/3 / LB = 1/3

option B is correct

QuantMadeEasy · Answer

DC = 2x ; BC = 3y

height and base of the triangle is 2y and 2x respectively
area of the triangle = (1/2) * 2y * 2x = 2xy

area of rectangle = 6 xy

ratio = 1/3

IMO B

SUNNYRHODE002 · Answer

Answer : C

Solution : In the picture below.

lacktutor · Answer

Please, find the attached file

Posted from my mobile device

Archit3110 · Answer

let side AB = 6 and AD = 4
area of ABCD = 24
and ∆ AED ; 3:4:5 ; area = 1/2 * 3*4 ; 6
and ∆ ABC = 1/2 * 6* 4 ; 12
so area of ∆ DOC = 24 -(12+6) ; 6 
ratio the ratio of the area of shaded triangle DFC and the area of the rectangle ABCD 
6/24 ; 1/4
IMO C

In the figure above, the quadrilateral ABCD is a rectangle and AE= EB. What is the ratio of the area of shaded triangle DFC and the area of the rectangle ABCD?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. Cannot be determined

vishumangal · Answer

Ans should be E as provided Info is no sufficient to ans the Question

Aviral1995 · Answer

Triangle AFE is similar to triangle CFD (BY AAA Similarity)
therefore AE/DC=1/2
--> (Height of triangle AFE) / (Height of triangle CFD) = 1/2
LET HEIGHT OF TRIANGLE AFE=b ; Then HEIGHT OF TRIANGLE DFC=2b
total length of AD=3b

ratio of area of DFC to area of rectangle abcd = 1/3

option B

Raxit85 · Answer

I don't know how to proceed but still ill try with scrappy approach.

Let me assume, AB = 4 and AD = 2, so AE =2 & EB = 2,
If i assume that the shaded region is triangle DEC, then, the area of DEC will be 1/2*4*2 = 4, (but actually it will be bit less than 4)
Area of ABCD rectangle = 4*2 = 8.

If i consider area of shaded region as 3.5 = 7/2,
then area of shaded region/area of rectangle = 7/2 divided by 8 = approx. 0.4, 
So, i'd guess 1/3 (0.33) as correct answer as it's nearest to 0.4.

Ans. is B

fauji · Answer

Used Visual approach, may not be right, However would like to submit my entry here as last few second question when time's almost gone:

- From the figure, it looks like slightly more than 1/4 and less than 1/2.

- Negated Option A,C, and D, Narrowed down to Option B and E.

IMO Option B!

hiranmay · Answer

In the figure above, the quadrilateral ABCD is a rectangle and AE= EB. What is the ratio of the area of shaded triangle DFC and the area of the rectangle ABCD?
A. 1/2
B. 1/3 -->  correct 
C. 1/4
D. 1/5
E. Cannot be determined

Solution :
let's say length of ABCD = l (=AB=C) & width of ABCD = w
height of DEF = h, so height of AEF =w-h
area of the rectangle ABCD = (area of ADE + area of ABC - area of AEF) + area of DFC
=> lw =   + (1/2)*l*h
=> 4lw = lw + 2lw - l(w-h) + 2lh= 2lw+3lh
=> 2lw=3lh
=> h /w= 2/3

So (area of shaded triangle DFC) /(area of the rectangle ABCD)=(1/2)*l*h / lw=h/2w=1/3

globaldesi · Answer

triangle AEF and traingle DFC are similar
DC = 2AE
thus the altitudes of corresponding triangles will also be of ratio 1:2 
hence the width is divided in 3 parts
height of triangle AEF = 1/3 (width of triangle) 
 =\frac{w}{3} 
height of triangle DFC = 2/3 (width of triangle) 
 =\frac{2w}{3} 
area of triangle DFC thus =  \frac{1}{2} * l *\frac{ 2}{3} w 
=  \frac{1}{3} lw 
thus ratio will be
 \frac{1}{3}* lw /lw 
= 1/3

AndreV · Answer

Triangle AEF ~ Triangle DFC

h2/a=h1/a. From this equation, h1=2h2

Then,

Area (ADF) = (2a)h1/2 = a(2h2) = 2ah2
Area(ABCD) = (2a)(h1+h2)=2a(2h2+h2)=2a(3h2)=6ah2

Therefore,
Area (ADF)/Area(ABCD)=2ah2/6ah2=1/3

Answer: 1/3.

samarpan.g28 · Answer

In the figure above, the quadrilateral ABCD is a rectangle and AE= EB.

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In the figure above, the quadrilateral ABCD is a rectangle and AE= EB.

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