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805+ (Hard)|   Geometry|      
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Bunuel
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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, 25 Mar 2019, 00:08
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30% (03:08) correct 70% (02:56) wrong based on 86 sessions

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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
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C
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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, 01 Oct 2019, 01:24
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Bunuel
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
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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, 01 Oct 2019, 05:12
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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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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, 03 Oct 2019, 00:33
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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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In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, 06 May 2021, 01:10
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If you know the theorem, you can solve the problem rather quickly. You can also prove it by extending the bottom parallel side of the trapezoid

(1st) Trapezoid Mid-Segment Theorem ——> states that when you draw a Line from the Mid-point of one of the non-equal sides to the Mid-point of the opposite non-equal side, that Line will be:

—Parallel to the 2 Parallel Sides (Here: AB parallel to DC parallel to FE)

—- and the length of this Mid Segment will equal the Arithmetic Mean of the lengths of the 2 parallel sides

—- further, since FE, is the Mid Segment of the trapezoid, the Perpendicular Height between parallel lines AB and DC will be cut in Half by parallel mid segment FE

This means each of the 2 split trapezoids will have Height = (h/2)

Let AB = X

Let DC = Y

Question asks for: What is (AB/DC) = (X/Y) = ?


FE, from mid point to mid point will equal = (X + Y)/2

Now, just set up the given information and solve for the ratio of (X/Y):


Area of trapezoid ABEF = (2) * (Area of Trapezoid DFEC)

(1/2) [AB + FE] (h/2) = (2) (1/2) [FE + DC] (h/2)

You can cancel (h/2) from each side and (2) (1/2) on the right side = 1

Substituting the variables we made up, we have:

(1/2) [ X + (X + Y / 2) ] = [ (X + Y / 2) + Y]

Confining like terms with a common DEN of 2 inside the parenthesis, we have:

(3X + Y) / 4 = (X + 3Y) / 2

6X + 2Y = 4X + 12Y

2X = 10Y

X = 5Y

(X/Y) = 5

Edit: Note, you do not need to assume that the trapezoid is an Isosceles trapezoid for the Mid Segment Theorem for Trapezoids to apply. It applies even if it is a regular Trapezoid.

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