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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