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# 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

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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 such that the ratio ofthe areasofthe two parts formed is 2:3 , ompute the length ofthe dividing line.

Guest Dec 28, 2014

#1
+80935
+10

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....!!!!

CPhill  Dec 28, 2014
Sort:

#1
+80935
+10

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

--------------------------------------------------------------------------------------------------------------

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....!!!!

CPhill  Dec 28, 2014
#2
+91435
+5

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.

Melody  Dec 29, 2014
#3
+80935
+5

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

CPhill  Dec 29, 2014
#4
+91435
0

okay , thanks Chris

Melody  Dec 30, 2014

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