Finding the perimiter of an irrecular pentagon

[jg]

I have a 5 acre piece of property that I wish to split into 5 individual 1 acre parcels. The 5 acre parcel 1s 660 feet in length and 330 feet in width. The first two parcels are 264'x165', and are located at the small base of the rectangle. I am asking for help getting the perimiters of the last 3 lots, one of which is a irregular pentagon. All of the one acre parcels must have aan area of 43,560 square feet each. Can anyone help me? Thanks,

John

 

[TchrWill]

I have a 5 acre piece of property that I wish to split into 5 individual 1 acre parcels. The 5 acre parcel 1s 660 feet in length and 330 feet in width. The first two parcels are 264'x165', and are located at the small base of the rectangle. I am asking for help getting the perimiters of the last 3 lots, one of which is a irregular pentagon. All of the one acre parcels must have aan area of 43,560 square feet each.

Draw rectangle ABCD, A uper left, B upper right, C lower right and D lower left.

AB = CD = 660

AD = BC = 330

Draw line EF parallel to, and 264 ft. from, side AD.

Draw line GH from the mid point of AD to the mid point of EF.

From B and C, locate two points along BA and CD respectively to points J on BA and K on CD.

Extend GH out to a point L closer to G than J and K are to A and D.

Draw lines JL and KL.

BCKLJ is your irregular pentagon.

You must play with the location of points J, L and K such tha area of pentagon BCKLJ is equal to 43, 560 sq. feet.

When you find such a solution, the areas of HEJL and HFKL will each equal 43, 560 sq. ft.

The perimeters of the las 3 lots are not estabished.

The problem is there are an infinite number of solutions depending on where points J and K are located.

If, by chance, you are seeking the minimum perimeter of the last 3 lots, then you can set up the equations of the 3 lot areas in terms of JB and KC and differentiate.

 

[Denis]

Using TchWill's labelling, if you "fix it" such that JB = KC, then the pentagon BCKLJ
has side BC = 330, with JL = KL and JB = KC.
This means the other 2 lots are identical.

Then (as example of one integer solution) if JB = KC = 88 and HL = 220,
JL and LK will equal 187, and the perimeters will all be 880.

There are 4 integer solutions; the other 3 have different perimeters:
(JB/KC , HL, perimeter EJLH/FKLH, perimeter BCKLJ)
22 154 968 924
60 192 912 888
88 220 880 880 (the example above)
106 238 866 888

 

[stapel]

Can anyone help me?

Even if we had all the geometric information (and, as the other tutors have mentioned, we do not), there is real-world information that we don't have (such as the validity of the measurements, the elevations, the location of legal landmarks, the zoning and set-back regulations for that area, etc).

To obtain a survey of the property, please hire a licensed professional. Your local "yellow pages" may have listings. And you might want to consider hiring a real-estate attorney regarding splitting the property, as smaller-sized lots are not always conforming.

Eliz.

 

[Denis]

> I have a 5 acre piece of property that I wish to split into 5 individual 1 acre
> parcels. The 5 acre parcel is 660 feet in length and 330 feet in width. The first
> two parcels are 264'x165', and are located at the small base of the rectangle.

WHY do you say "that I wish to split"?
Since you have 2 already located, seems you can't "wish" no more!

> I am asking for help getting the perimiters of the last 3 lots, one of which is a
> irregular pentagon. All of the one acre parcels must have an area of 43,560
> square feet each.

WHY is one an irregular pentagon: is that a decision you took?

You seem to be left with a 330 by 396 rectangle (removing the two 264 by 165),
and want to "fence" that such that you have at least one parcel that is in shape
of an irregular pentagon, and don't care about the shape of the other two.

Gee John, can you not at least place some corner posts on the sides of that
330 by 396 rectangle, AND TELL US where they are?

 

[TchrWill]

I have a 5 acre piece of property that I wish to split into 5 individual 1 acre parcels. The 5 acre parcel 1s 660 feet in length and 330 feet in width. The first two parcels are 264'x165', and are located at the small base of the rectangle. I am asking for help getting the perimiters of the last 3 lots, one of which is a irregular pentagon. All of the one acre parcels must have aan area of 43,560 square feet each. Can anyone help me? Thanks,

Draw rectangle ABCD, A uper left, B upper right, C lower right and D lower left.

AB = CD = 660

AD = BC = 330

Draw line EF parallel to, and 264 ft. from, side AD.

Draw line GH from the mid point of AD to the mid point of EF.

From B and C, locate two points along BA and CD respectively to points J on BA and K on CD.

Extend GH out to a point L closer to G than J and K are to A and D.

Draw lines JL and KL.

BCKLJ is your irregular pentagon.

You must play with the location of points J, L and K such tha area of pentagon BCKLJ is equal to 43, 560 sq. feet.

When you find such a solution, the areas of HEJL and HFKL will each equal 43, 560 sq. ft.

The perimeters of the las 3 lots are not estabished.

The problem is there are an infinite number of solutions depending on where points J and K are located.

If, by chance, you are seeking the minimum perimeter of the last 3 lots, then you can set up the equations of the 3 lot areas in terms of JB and KC and differentiate.

Letting h = the altitude of triangle JKL from L to JK, x = JB = KC, y = HL and z = JL =LK:

h = 2264 - 2x
y = 396 - h -- x
z = (4x^2 - 528x + 96,921)^(1/2)

The total perimeter of the three plots including the irregular pentagon is P = EB += BC + CF + FE + JL + LK + HL.

As x varies, EB + BC + CF + FE remain constant.

Therefore to minimize the perimeter of these 3 pots, we need only find the minimum sum of JL += LK + HL.

The total length of JL + LK + HL is L = the sum of y + 2z which comes out to L = x + 132 + (4x^2 - 528x + 96,961)^(1/2).

Take the 1st derivitive, set the result equal to zero, and solve for x.