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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,

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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,  [#permalink]

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New post 25 Mar 2019, 00:08
1
7
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:48) correct 69% (02:47) wrong based on 48 sessions

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Location: India
Concentration: Entrepreneurship, Marketing
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Re: In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,  [#permalink]

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New post 01 Oct 2019, 01:24
1
Bunuel wrote:
In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F as the midpoint of DA. The area of ABEF is twice the area of FECD. What is AB/DC ?

(A) 2
(B) 3
(C) 5
(D) 6
(E) 8


For ease of doing question, Assume ABCD as an ISOSCELES triangle
—> Trapezoids ABEF & FECD are also ISOSCELES and the heights between these trapezoids is also same (assume as h)

Let AB = x, EF = y, DC = z,
Given, Area(ABEF) = 2*Area(FECD)
—> 1/2*h*(x + y) = 2*1/2*h*(y + z)
—> x + y = 2y + 2z
—> y = x - 2z ...... (1)

Also, Area(ABCD) = Area(ABEF) + Area(FECD) = 3*Area(FECD)
—> 1/2*2h*(x + z) = 3*1/2*h*(y + z)
—> 2x + 2z = 3y + 3z
—> y = (2x - z)/3 ....... (2)

From (1) & (2),
x - 2z = (2x - z)/3
—> 3x - 6z = 2x - z
—> x = 5z
—> x/z = 5

So, AB/DC = x/z = 5

IMO Option C

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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,  [#permalink]

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New post 01 Oct 2019, 05:12
For a trapezoid the Area= (a +b)/2

a and b the sides parallel

Let m be the line that joins the mid points E and F

m*h = (a+b)/2 *h
m =(a+b)/2

Given
ARea of ABEF = 2 Area of FEDC

Area of the parallelogram ABCD = ARea of ABEF + Area of FEDC
(a+b)/2 *h = 3 *Area of FEDC
(a+b)/2 *h = 3 *(b+m)/2 *h/2 (Since the triangles will be made parallel)
(a+b)= 3/2(b +(a+b)/2)
a+b = 3/2 (3b+a)/2
a+b =9b/4 + 3a/4
a/4 = 5b/4
a/b= 5

Ans C

Or

(a+m)/2 *h /2 = 2(b+m)/2 *h/2
a-2b= m

(a+b)/2 *h = 3(b+d)/2 *h/2
2a+2b = 3a-3b
5b= a
a/b = 5

C
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Re: In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,  [#permalink]

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New post 03 Oct 2019, 00:33
ar(ABEF) = 2 ar(FECD)

let AB= a
FE= m
DC= b

1). ar(ABEF) = 2ar(FECD) ar of trapezoid= (a+b)/2 * h
(a+m)/2 * h/2 = 2[ (b+m)/2 * h/2] the mid point divides the height into 2 equal parts
a+m = 2b+ 2m

m= a-2b


2). ar(ABCD) = ar(ABEF) + ar(FECD)

(a+b)/2 *h = 3 ar(FECD)

(a+b)/2 * h= 3[ (b+m)/2 * h/2]

(a+b) = 3[ b +a -2b]/2 putting the value of m as calculated above

a+b = 3/2(a-b)

2a+2b=3a - 3b

a/b =5

the awnser is C :) :thumbup:
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Re: In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC,   [#permalink] 03 Oct 2019, 00:33
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