Calli H.
asked • 12/05/14Where can they position the new fence so the areas of the two landowner's property remain the same?
Two landowners, Mr Brown and Mrs Lopez, agree to put up a new fence between their properties. Rather than replacing the fence as it is now (with a third fence post), they want to put in a new fence that is straight, thus requiring only the two end posts. Where can they position the new fence so the areas of the two landowner's property remain the same?
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1 Expert Answer
Nevine M. answered • 12/07/14
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Draw two vertical lines on your diagram. One line goes through the 2 end points of the fence, and the other line is parallel to the first one, but it goes through the single middle point of the existing fence. You should have a rectangle with 3 triangles in it (a big one on Mrs. Lopez's side, and 2 little ones on Mr. Brown's side). Let's call the horizontal width of the rectangle a, and the vertical length of the rectangle b.
If you look at the triangles Mrs Lopez's large triangle you can calculate the area as being (1/2)ab.
The area of Mr. Lopez's 2 small triangles is 2(1/2)(1/2)a(1/2)b) which simplifies to (1/2)ab
The areas are the same, so they both have the same amount of land. You should therefore position the straight fence in the center of a (your rectangular width) to give each of them half of the rectangle.
Annie O.
But if you draw a line parallel to the first one, then wouldn't it be a trapezoid, not a rectangle? And it's not necessarily a regular trapezoid...
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01/07/15
Nevine M.
Oh I see, I misinterpreted the diagram. If I understand it correctly now it is a little bit more complicated but here is what I believe needs to be done.
Draw the 2 vertical (parallel) lines. You will have 3 triangles.
x = length of vertical line on Mr. L's side
y =length of vertical line on Mr. B's side
Extend a horizontal line (perpendicular to the vertical lines that you drew) from x to y, let's call this "h"
Now draw a third vertical line somewhere in between x and y. Let's call its length "k"
Horizontally, from x to k is measure "a", and then from k to y is measure (h-a)
Initially, we have 3 triangles.
The area of Mr. B's 2 triangles is:
A (belonging to Mr B) = (1/2)(y)(h)
The area of Mr. L's 1 large triangle is:
A (belonging to Mr. L = (1/2)(x)(h)
With the new format of a straight line, this will split the land into 2 trapezoids.
Mr. L's trapezoid will have bases x and k and the height will be a.
Mr. B's trapezoid will have bases k and x and the height will be (h-a)
Using the equation for area of a trapezoid A=(1/2)(b1+b2)(h) We can set up equations for the areas of each new trapezoid.
Mr. B's area
A (belonging to Mr B) = (1/2)(k+y)(h-a)
Mr. L's area
A (belonging to Mr L) = (1/2)(x+k)(a)
Set Mr. B's triangular area equal to the trapezoidal area and do the same for Mr. L to get a system of equations.
(1/2)(y)(h) = (1/2)(k+y)(h-a) Mr. B Eqn 1
(1/2)(x)(h) = (1/2)(x+k)(a) Mr. L Eqn 2
The (1/2) cancels out of both equations, leaving you with:
(y)(h) = (k+y)(h-a) Eqn 1
(x)(h) = (x+k)(a) Eqn 2
We want to solve for "a". This will give us the position of the new fence post.
Does that help?
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01/09/15
Nevine M.
There is a typo above when describing Mr B's new area.
With the new format of a straight line, this will split the land into 2 trapezoids.
Mr. L's trapezoid will have bases x and k and the height will be a.
Mr. L's trapezoid will have bases x and k and the height will be a.
Mr. B's trapezoid will have bases k and y and the height will be (h-a).
Report
01/09/15
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John G.
12/05/14