Please help me to solve this question........... 4AB=5CD, OB=3BD, and AB//CD. That is, AB is 1.25 times longer than CD; OB is three times longer than BD; and AB and CD are parallel. If the area of a triangle ACE is 1cm then how big is the area of ECDB?

copy and paste the url to see the diagram : - http://a1.sphotos.ak.fbcdn.net/hphotos-ak-snc7/398...

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*IF* I can assume that OC is perpendicular to AB, then here is the solution. However, this is not stated in the problem.

First, notice that ECDB is a trapezoid. The formula for the area of a trapezoid is

A = (h/2)(b1 + b2)

Okay, here we go....

Note that EOB and COD are similar triangles (you can verify this).

OD = OB + BD

OD = 3BD + BD <----- Given: OB = 3(BD)

OD = 4BD

And thus the scale factor of COD to EOB is 4BD / 3BD = 4/3. Triangle COD is 4/3 times larger than EOD. This means CD = (4/3)(BE), and thus

b1 + b2 = BE + (4/3)(BE) = (7/3)(BE)

Halfway there... now let's figure out the height of ECDB.

AB = (5/4)(CD) <----- Given

AB = (5/4)[(4/3)(BE)]

AB = (5/3)(BE)

AB = (3/3 + 2/3)(BE)

AB = BE + (2/3)(BE)

AB - BE = (2/3)(BE)

AE = (2/3)(BE).

The area of triangle AEC is 1 cm^2

A = (1/2)bh

1 = (1/2)(AE)(CE) <----- This is where I assume AB & OC are perpendicular.

2 = (AE)(CE)

2 = (2/3)(BE)(CE)

3/(BE) = CE

Now you can determine the area of the trapezoid.

A = (h/2)(b1 + b2)

A = (1/2) [3/(BE)] [(7/3)(BE)]

A = 7/2 cm^2

But again, take that with a grain of salt because it was not explicitly specified in the problem statement, nor is there any indication in the diagram, that OE and AB are perpendicular.