Math Wizzard Needed!

[Guest ncm2169]

No, this isn't about how to measure cock. ;-)

 

My math skills are rusty. I'm looking at a piece of property and I'd like to know the square acreage. The lot is irregular with the following dimensions: 159ft X 157ft X 110ft X 133ft. Can anyone calculate the square acreage for me?

[+ Tom Isern]

NCM...

 

This is a tough one, but the formula is found here:

 

http://www.onlineconversion.com/forum/forum_1090204749.htm

 

Good luck!

[Guest ncm2169]

Thanks Tom. Where's the answer? ;-)

[badassbadboy]

If I am correct one acre is 330' X 350' so what you may have there is a half an acre.

 

Hope that helps.

[poiuyt]

You haven't enough information to determine the area. Four sides of given lengths can make a variety of four-sided figures.

 

If you know one of the diagonals, then you have two triangles, and can then get the area (since three lengths can make only one triangle).

 

Suppose you have triangle with the sides A, B and C.

 

Let S = (A+B+C)/2 (or half the perimeter).

 

Let T be the result of multiplying S, (S-A), (S-B) and (S-C).

 

Then the area is the square root of T.

 

(Heron's formula)

[Guest zipperzone]

>If I am correct one acre is 330' X 350' so what you may have

>there is a half an acre.

>

>Hope that helps.

 

Sorry Tony, but you're wrong - 330' X 350' = 115,500 sq. ft. What you have described is 2.65 acres

 

An acre is only 43,560 sq. ft. Trust me on this!

[flguy]

NCM, I answered you elsewhere but let me try again. This is a little better than a guess but I am not a surveyor. Appears to be about 19,208 square feet or .44 acres. Second opinion is desired.

[+ Tom Isern]

As someone else has already pointed out, you didn't give enough information for an answer. The formula is on the website I gave you, though...

[mercury40]

close, the correct answer is 19,502.5 sq ft.

[Zippy]

>My math skills are rusty. I'm looking at a piece of property

>and I'd like to know the square acreage. The lot is irregular

>with the following dimensions: 159ft X 157ft X 110ft X

>133ft. Can anyone calculate the square acreage for me?

 

 

As another poster already pointed out, you have not provided enough information in order for anyone to compute the area or acreage. So, everybody, stop trying to spout out what the "correct" answer is. You are all wrong until NCM tells us whether or not there are any perpendicular sides to the lot or provides us with one of the diagonal measurements of the quadrilateral. Without this information, the problem is unsolvable.

 

 

To quote the other poster:

 

"You haven't enough information to determine the area. Four sides of given lengths can make a variety of four-sided figures.

 

If you know one of the diagonals, then you have two triangles, and can then get the area (since three lengths can make only one triangle).

 

Suppose you have triangle with the sides A, B and C.

 

Let S = (A+B+C)/2 (or half the perimeter).

 

Let T be the result of multiplying S, (S-A), (S-B) and (S-C).

 

Then the area is the square root of T.

 

(Heron's formula)"

[flguy]

I don't believe you need a perpendicular triangle to figure out the area. Take ANY triangle and a mirror image of that triangle and put together and they make a rectangle, albeit off kilter a bit. Length time width equals area and divide by two and that gives you the area of the triangle. Try it and I think you will see it works.

[mercury40]

flguy, no need to be a rocket scientist to figure this one out, you've given one way of arriving at the answer... If his property had been 60' x 100', it would be easy. Lenght times width would equal 6,000 sq ft. However, it's an odd shape. So you have to take the average length and width to arrive at the square footage and then divide it into the sq ft of an acre. Good luck to all you so called "WIZZARDS" with formulas! You're reading way too much into this simple 5th grade math problem.

[Zippy]

>I don't believe you need a perpendicular triangle to figure

>out the area.

 

I didn't say you need a "perpendicular triangle". In fact, a 3-sided figure in which two sides are perpendicular would be called a right triangle, not a perpendicular triangle. What I said was that we need to know whether or not any of the sides of the quardrilateral were perpendicular to eachother. Then, we could find the area of the figure.

 

 

> Take ANY triangle and a mirror image of that

>triangle and put together and they make a rectangle, albeit

>off kilter a bit. Length time width equals area and divide by

>two and that gives you the area of the triangle. Try it and I

>think you will see it works.

 

 

"albeit off kilter a bit"? That wouldn't be a rectangle anymore. Rectangles have right angles, or perpendicular sides. If it is "off kilter" even a little bit, you can't just multiply the length times the width, because quadrilaterals that do not have perpendicular sides do not "lengths" and "widths" as defined this way.

[Zippy]

>flguy, no need to be a rocket scientist to figure this one

>out, you've given one way of arriving at the answer... If his

>property had been 60' x 100', it would be easy. Lenght times

>width would equal 6,000 sq ft. However, it's an odd shape. So

>you have to take the average length and width to arrive at the

>square footage and then divide it into the sq ft of an acre.

>Good luck to all you so called "WIZZARDS" with formulas!

>You're reading way too much into this simple 5th grade math

>problem.

 

 

You may not need to be a rocket scientist, but you need a lot more Geometry than that of a 5th grader. Your idea of just multiplying the "length" times the "width" to obtain an answer of 6,000 sq. ft. only works if the original figure is a rectangle. Which, clearly, this figure is not because the opposite sides would have to be equal in length, and they are not.

 

And, sorry to say, but your idea of just averaging the given numbers to try to come up with an approximate length and width will not work, either, unless some of the given measures are known to be perpendicular to one another.

 

Maybe you should leave the complicated stuff up to the "wizzards"...

[+ ArVaGuy]

Where were you guys when I was flunking geometry in high school?

 

ArlingtonVaGuy

[mercury40]

RE: Zippy is not one of our Math Wizzards!

 

LOL Zippy, It's time for you to go back to school. The formula used by flguy is absolutely correct. It's better you post about things you know about, obviously, Math is not one of them. :-)

[+ Hoover42]

I've written a Windows program to calculate the area.

 

http://mywebpage.netscape.com/scudder1955/AreaCalc.zip

 

...Hoover

 

p.s. The program requires the lengths of the four sides plus one diagonal length in feet, so it works only with concave quadrilaterals (as Boston Guy notes below).

[+ Tom Isern]

From the website I mentioned earlier:

 

Re: irregular yard deminsions to acres

by Robert Fogt on 07/30/04 at 19:37:26

 

The only way to calculate the area of an irregular polygon, is to also know either one of the angles, or the length of one diagnal.

 

Measure the length of one of the diagnals, that will be enough to figure it out.

 

What we do, is split the area into two seperate triangles, then calculate the area of each triangle seperately using Heron's formula.

 

Heron's formula is:

Area = sqrt(s(s-a)(s-b)(s-c))

Where s = (a+b+c)/2

 

 

Here's an example:

 

The first triangle, a = 230, b = 243, c = 216.5

Solve for s:

s = ( 230 + 243 + 216.5) / 2 = 344.75

Now enter into Heron's formula:

Area = sqrt(344.75(344.75 - 230)(344.75 - 243)(344.75 - 216.5)) = 22720.8398 square feet

 

The second triangle, a = 186, b = 296, c = 216.5

Solve for s:

s = (186 + 296 + 216.5) / 2 = 349.25

Now enter into Heron's formula:

Area = sqrt(349.25(349.25 - 186)(349.25 - 296)(349.25 - 216.5)) = 20075.7544 square feet

 

Adding the area of the two triangles together for the total area of the yard:

22720.8398 + 20075.7544 = 42796.5942 square feet.

 

Hope this is helpful...

[Boston Guy]

As others have stated above, you haven't provided enough information to answer the question. Any estimate provided based on the lengths of the sides alone may be off by quite a bit. If you are seeking the answer in order to make a decision on a piece of real estate, you are well-advised to go back to the property and measure one of the diagonals.

 

Just to get some terms out of the way:

 

* Polygon: a figure on a plane that is comprised of line segments that are joined together in such a way that the figure is closed (meaning you can walk all around the perimeter and get back to your starting point). Polygons also have the property that the sides of the property don't cross each other and only two lines meet at each intersection.

 

* Any four sided polygon is called a quadrilateral (literally, "four-sided").

 

* Any quadrilateral with two parallel sides is called a trapezoid.

 

* Any trapezoid with both sets of opposite sides parallel to each other is a parallelogram.

 

* Any parallelogram with four equal angles (necessarily 90 degrees each) is a rectangle.

 

* Any rectangle with sides of equal length is a square.

 

If you take all quadrilaterals with a given perimeter, the square with that perimeter will enclose the largest area. (This is called the isoperimetric problem for quadrilaterals.)

 

For example, take two pieces of land:

 

a. Parcel A has four sides, each 100 feet long. It contains 10,000 square feet of land.

 

b. Parcel B is a rectangle. Two of the sides are 20 feet long and the other two are 180 feet long. It has the same perimeter as Parcel A -- 400 feet -- but contains only 3,600 square feet of land. If you attempt to estimate the area of Parcel B by averaging the sides, you will obtain (180+20)/2 = 100 for the average length of each side. 100 x 100 = 10,000 square feet. Your quick estimate will inform you that the property is roughly three times larger than it really is.

 

All parcels of land that have a perimeter of 400 feet but are not squares will contain less than 10,000 square feet. The only question is how much less.

 

When we think of shapes, we tend to think of convex shapes. However, when purchasing land sight unseen, beware of the concave shape.

 

A convex shape is one where if you draw a straight line connecting any two intersections, that connecting line will be entirely within the shape. For example, if you a draw a line connecting any two corners of a square, the line will be inside the square (or be one of the boundaries). You cannot draw a straight line connecting two of the corners of a square and have that line be outside of the square.

 

Not so with concave shapes.

 

Take four toothpicks. Break off a quarter or so of two of them. Take the two longer ones and connect them so they look like a kind of wide-open "V". Take the short ones and connect them to the ends of the first two and make the second two connect inside the "V".

 

What you will have just done is outline a kind of fat "V" shape with four toothpicks. You will also have created a concave shape. And you can imagine a parcel of land this shape that has a perimeter of 400 feet and an area far less than a square with the same perimeter.

 

Creating a concave quadrilateral with a given perimeter is the easiest way to disprove the nonsense that you can somehow calculate the area of a piece of land by knowing only the length of its sides. You cannot, and no manner of averaging will yield a correct answer. In fact, no manner of averaging will yield an answer that you can count on as being even close to correct.

 

If you walk the land and know that it's roughly a square, then you can make estimates about the relationships of the sides to each other and come up with an estimate that may serve your purpose, which you have not stated here. But if you have simply been given the lengths of the four sides of a piece of land you have never seen, you could easily be talking about a convex quadrilateral that is significantly smaller than you might imagine.

 

If there is money involved and it's important to you, measure a diagonal.

 

BG

 

 

ps: After posting this and walking away, I realized that if one is purchasing a piece of land, sight-unseen, from an unknown seller, it isn't enough to measure one of the diagonals; you need to know the length of both of the diagonals.

 

In the case of our "fat V"-shaped concave quadrilateral, above, one of the diagonals is outside of the polygon -- namely, the diagonal connecting the two top points of the "V". This line is, in fact, a diagonal and it has a true length. But it's not inside the shape and if you were only supplied that one measure, you'd think you were dealing with a convex shape and, consequently, a much larger piece of land. Only knowing the length of the other diagonal, which will be very, very small in the case of our "fat V", will bring the truth to light.

 

And, in case anyone is wondering is this is real-world stuff, all you have to do is go look at property boundaries wherever they are registered in your locality. Many pieces of land have very irregular shapes and some of them are indeed concave.

[Zippy]

RE: Zippy is not one of our Math Wizzards!

 

>LOL Zippy, It's time for you to go back to school. The

>formula used by flguy is absolutely correct. It's

>better you post about things you know about, obviously, Math

>is not one of them. :-)

 

 

Ok, moron, why is it so difficult for you to read all the other info posted by the other people on this thread? Or, click on the link provided in the very first reply to the question for more information on this topic. The answer to this problem is not as easy as you and flguy seem to think it is. I'm sorry that your ability to grasp such concepts is above you, but it clearly is.

[Zippy]

Its refreshing to see that there are, indeed, other people out there in the World who actually do understand the basic concepts of Geometry. Great job, BG!

[+ deej]

Showoff.

[mercury40]

RE: Zippy is not one of our Math Wizzards!

 

>Ok, moron, why is it so difficult for you to read all the

>other info posted by the other people on this thread? Or,

>click on the link provided in the very first reply to the

>question for more information on this topic. The answer to

>this problem is not as easy as you and flguy seem to think it

>is. I'm sorry that your ability to grasp such concepts is

>above you, but it clearly is.

 

LOL! Zippy, the true sign of a LOSER is when you have to resort to name calling. That's taught in debate class 101... As I said before, stick to things you know something about and leave the Math to true "Wizzards" like me and flguy. :-)

[bigguyinpasadena]

Y'all need to come visit me-I live 4 blocks from Cal-Tech you should hear some of the conversations in the nearby coffee shops(i.e.Pie and Burger)these boys have no idea how adorable they are-and soooooooo smart!

[Zippy]

RE: Zippy is not one of our Math Wizzards!

 

If you weren't so stupid, you would actually be funny. Evidently, besides not knowing anything about mathematics, you can't read, either. Try stumbling over all the big words in BG's post for the next few hours, and then maybe, just maybe, you might come close to understanding how wrong you are. But, I doubt it.