A trapezoid 6m in altitude have bases of 12 m and 18m, respectively, If the it is divided into two parts by a line parallel to the bases suc

The area of the whole trapezoid is(1/2)(18 + 12)(6)= (15)(6) =   90m^2

Let the upper base = 12 and let the lower base = 18

And since a line divides the area in a ratio of 2:3, let the area of the upper trapezoid = 36 and let the area of the lower trapezoid = 54

Let 12 + 2x be the length of the dividing line and let y be the height of the lower trapezoid...so we have

(1/2)(12 + 2x + 18)(y) = 54

(30 + 2x) y = 108

2(15 + x)y = 108

(15 +x)y = 54

y = 54/(15 + x)

And the area of the upper trapezoid can be represented by

(1/2)(12 + 12 + 2x)(6-y) = 36

(24 + 2x)(6-y) = 72  .....  factor out a "2"

(12 + x)(6-y) = 36

6 - y = 36/(12 + x)

y = 6 - 36/(12 + x)

And equating "y's"     ....we have

54/(15 + x) = 6 - 36/(12 + x)   and soving for x, we have

x = 3(√6 - 2) = about 1.3484692283495343m....so 2x = about 2.6969384566990686m

So...the length of the dividing line is about (12 + 2x)m = 14.6969384566990686m

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Note that this problem would probably have a different answer if we kept the orientation of the bases the same and considered the upper trapezoid to contain 3/5ths of the area and the lower one to contain 2/5ths of the area. I'll let the adventurous types work through that one....!!!!

Thanks for that great answer Chris 

The length of the dividing line would definitely be different (longer) is the areas were divided the other way around :)

I am curious as to why you made the dividing line 12+2x units.  Did that make the maths easier?    

I would just have called it x

I have not worked the maths through - your way may be much easier.   

Yeah, Melody, using "x" for the length probably would have made it easier.....I just didn't think to do it that way.....!!!

okay , thanks Chris