Math Wizzard Needed!

[Lookin]

Flguy is right, if you assume the property is pretty close to a rectangle.

 

In that case, it has an average depth of 158 feet (159 plus 157, divided by 2), and an average width of 121.5 feet (110 plus 133, divided by 2). That's equal to 19,197 square feet, or .44 acre (19,197 divided by 43,560).

 

If it's not pretty close to a rectangle, then all bets are off. Imagine a skinny diamond-shaped property with the same side-length measurements as your property. It would have a much smaller area, and could even be close to zero. That's where you'd have to get into angles and stuff. But that would be a pretty useless piece of property. The most reasonable assumption is that it's close to a rectangle, and close to four-tenths of an acre.

 

Plenty big enough to lay your hat and a few friends. (One of Dorothy Parker's, not mine.)

[flguy]

Zippy, Try this. Take any 4 sided shape, just like the example we have been given. Divide it into 2 triangle with two known sides each. Take triangle one and multiply the two known sides together and then divide by two. Take the second triangle and do the same. Take the sum of the those two and you will arrive at the square footage of the property. Then divide by 43,560 to arrive at the percentage of acreage for that given plot. The only time this can not be used is when there is a curved line somewhere on the property. Try it before you condemn everyone for not being as bright as you.

[+ deej]

Of all the bitchslap fights we've seen around here over the years, a fight over algebra and geometry has to take the cake as the funniest one yet.

 

Can we go back to size queens arguing over who's bigger?

[Boston Guy]

Allen,

 

You are absolutely correct ... and in absolute agreement with everyone else who is saying that you need at least one diagonal to make even a good guess.

 

If you know enough about the shape of the property to be able to divide it into two triangles of known size, you're in business. Unfortunately, with only the lengths of four sides, we cannot come up with those triangles: the number of triangles that can be divised such that, when you put them together they form a piece of property with sides equal to the lengths we have been given, is very large. (Actually, I think the number of such pairs of triangles is infinite but I'm not willing to stand on that horse right now.)

 

One diagonal will reduce the number of such pairs to two -- making one convex shape and one concave shape. Having the other diagonal will reduce it to one pair and firmly define the size and shape of the entire property.

 

Without the diagonals, all one can do is guess that the property is likely to be somewhat rectangular, since most jurisdictions prefer to avoid truly odd-shaped properties. But that's only a guess and I sure wouldn't put my money into property based on such a guess.

 

Regards,

BG

[Boston Guy]

LOL... math is our friend. :-)

 

BG

[Boston Guy]

>Of all the bitchslap fights we've seen around here over the

>years, a fight over algebra and geometry has to take the cake

>as the funniest one yet.

>

 

Well, to my memory, it's the first thread centered on geometry in M4M's history. Sometimes it seems like every topic has been around before. It's nice to have an occasional change of pace! :+

 

BG

[Zippy]

>Zippy, Try this. Take any 4 sided shape, just like the

>example we have been given. Divide it into 2 triangle with

>two known sides each. Take triangle one and multiply the two

>known sides together and then divide by two. Take the second

>triangle and do the same. Take the sum of the those two and

>you will arrive at the square footage of the property. Then

>divide by 43,560 to arrive at the percentage of acreage for

>that given plot. The only time this can not be used is when

>there is a curved line somewhere on the property. Try it

>before you condemn everyone for not being as bright as you.

 

 

Wow, some people are really hard headed, huh? I am not condemning anyone; I am just trying to tell the originator of the thread, NCM, not listen to people who are telling him wrong information.

 

The process you are using to find the area of the figure only works, and I repeat, only works IF the figure has pairs of perpendicular sides. Which, NCM has not told us it does, and based on the lengths given, cannot possibly be true.

 

You may be thinking of the formula of (base)(height)/2 for a parallelogram, but even in a parallelogram, the height must be computed by using a perpendicular line drawn from the base to the opposite side. This height is often calculated using trigonometry, and in which case, we would need one of the angle measures between the sides of the original figure.

 

Hence, this is why both the website that Tom first posted, and the eloquent post made by BG each say that either some angle measures OR the length of one (or both) of the diagonals of the original figure are needed in order to compute the actual answer.

 

Do you really think that both of these sources, along with what I am saying are all wrong? Really, do you? Yeah, you probably do....

 

Ok, just for kicks, let's suppose that your idea works. Well, which sides do we multiply together first? Are you just going by the order that NCM wrote them down? Because, if you switch them around, you get a totally different answer:

 

(159)(157)/2 + (133)(110)/2 = 19,796.5

(159)(133)/2 + (157)(110)/2 = 19,208.5

(159)(110)/2 + (157)(133)/2 = 19,185.5

 

 

All of these answers are wrong, because as I said, this would only work if the figure contained all right angles, which it does not. Plus, just the fact that we keep getting different answers when changing the order of the numbers should be a hint that something is amiss.

 

Here are some more websites that may, or may not convince you further:

http://mathforum.org/library/drmath/view/63876.html

http://mathforum.org/library/drmath/view/54868.html

http://mathforum.org/library/drmath/view/52053.html

 

The first two are really good answers to problems just like the one that we are currently discussing, and I urge NCM to check them out if he really wants to find the answer to his problem.

 

 

 

However, since I am having so much fun with all of this math, and because I enjoy a good problem, lets investigate this problem further, shall we? Even if there are some perpendicular sides in the figure, without knowing which ones form right angles, we still can't come up with an answer.

 

If we assume that the side lengths are actually in the order that NCM wrote them down, 159, 157, 110, and 133, and that there is at least ONE right angle in the figure, then we can find the following: (Using the ideas of Pythagorean and Heron's theorems)

 

If there were a right angle between the 159 and the 157 length sides, then you would get an answer of 17,787.7

 

If there were a right angle between 159 and the 133 length sides, then you would get an answer of 19,066.2

 

If there were a right angle between 159 and the 133 length sides, then you would get an answer of 19,093.5

 

If there were a right angle between 133 and the 110 length sides, then you would get an answer of 18,734.8

 

 

Now, lets suppose that there is an 80-degree angle between the 159 and 157 length sides. Then, using the law of cosines, the diagonal squared would equal 157 squared + 159 squared - 2(157)(159)Cos(80), or 203.1. Then, using Heron's formula for the two triangles created with side lengths of 159, 157 and diagonal 203.1 and the other one with side lengths of 110, 133 and the diagonal of 203.1, the area would be: 19,021.9

 

But, if the angle between side lengths 159 and 157 were only 70 degrees, then using the same methods of law of cosines and Heron's formula, we would find the diagonal length to be 181.3 and the total area answer to be: 19,004.5

 

And, if the lot is really irregular, and the angle between the side lengths of 159 and 157 were something small like 15 degrees, then using the same methods of law of cosines and Heron's formula, we would find the diagonal length to be 41.3 and the total area answer to be: 11,998.9

 

Of course, since the sum of the lengths of any two sides of a triangle always have to add up to more than the length of the third side, it would be impossible for the figure to have a small angle like 10 degrees or less between sides 159 and 157.... but, we all know that, right?

 

 

Anyway.... I'm sure that the people that haven't been able to follow the information provided so far and still think that they can come up with the exact answer based on the original information given, will not believe anything that I have written, but oh well, I hope that NCM does.

[Zippy]

Sadly, BG, I think we may be wasting our breath, or rather our fingers on this one....

[Barry]

I am totally enjoying this. :7

[Zippy]

Hee hee, now is it that hottie boys with smokin' bodies and big dicks are our friends and math is our plaything, or is it the other way around....

 

 

Speaking of hottie play things and math, did you know that gay sex/relationships are actually mathematically correct, while str8 sex/relationships are flawed?

 

In gay sex/relationships, 1 + 1 = 2 always.

 

But, in str8 sex/relationships, 1 + 1 = 3, or 4 or 5, or... depending on just how many times he knocks her up.

 

So, since math is always right, gay sex is correct!

[+ Hoover42]

Hey, hasn't anybody tried my nifty utility yet. I guess you guys must all be Mac users.

 

...Hoover

[Boston Guy]

Nerd! }(

 

BG

[Guest FallenAngel]

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

 

I am not a mathematician and I do not play one on TV or on message center boards. But, if I was to comment on the above quote, I would simply say this.......

 

If there is money involved and it's important to you....HIRE A PROFESSIONAL!:7

[Guest ncm2169]

Hmmmm ... never had a new topic here, so I'm overwhelmed.

 

Sorry guys, I haven't the ability to determine exact angles yet. But I SINCERELY appreciate all the suggestions! :* :*

[+ Tom Isern]

Perhaps you need professional help...consult your real estate agent?

[jawjateck]

After reading some of the responses, note I said "some" of the responses, I'm impressed. I have never read such articulate discussions of a technical nature on this board. Some of you really can use that bigger head of yours.

 

Did you know people do not use more than 2% of their mental capacity? We apparently transferred the other 98% down to that little head. Looks like some of you guys are trying to reclaim those unused resources. This was a nice diversion and stimulated my thinking, but I'm still in that sub 2% crowd, so let's get back to the sex talk!!! }(

[flguy]

Tom, Maybe some of the people giving good advice on this subject are Realtors and professional also. Please try the following, you do NOT need a mile long formula, you do NOT need to know the length of any diagonal line or any of the other STUFF some are advocating. And to those who THINK they know, why have they never proved themselves with an answer???

 

Step 1: Draw a sample of the lot, does not need to be to scale. With the left side at 159 feet, the back at 157, the right side 110 and the front 133.

 

Step 2: (saying the rear of the lot is North) Draw a diagonal line from the NW corner to the SE corner. Do you now agree we have 2 triangles with two known measurements on two sides each? And for each triangle the measurements we have are the BASE and the HEIGHT?

 

Step 3: The formula for finding the area of a triangle is

Base times Height divided by 2 or

1/2(bh)=area

 

(159 X 133)/2=10,573.5 Square Feet

(157 X 110)/2= 8,635.0 Square Feet

 

10,574 plus 8,635 = 19,208 Sq Ft

 

Divide by 43,560 (number of square feet per acre)

 

EQUALS .44 acres.

 

If anyone would like to disprove this I would be happy to listen. But please do the math first with your fancy equations and make sure the answer isn't the same.

 

Any more questions???????

[Boston Guy]

Alan,

 

The formula you are using will not work in the way that you are suggesting. You are correct in saying that you can calculate the area of a triangle by using the formula b*h/2. However, the trick is in finding the right b and the right h.

 

You are assuming that the base and height of the triangles will be equal to the two sides of the triangle. This will only be true for a right triangle, where the angle between those two sides is 90 degrees.

 

We can see that it works in that case since a right triangle is half a rectangle and the formula for the area of a rectangle is width * length. Taking half of that gives the area for each of the right triangles that are constructed by drawing one of the rectangle's two diagonals.

 

Unfortunately, you cannot use the sides for b and h when the triangle is not a right triangle. Since the sides are all of different lengths here, at least one of triangles must not be a right triangle. (The other one might be or might not be; we don't have that information.)

 

When a triangle is not a right triangle, you can indeed use any one of it's sides as its base (b). But you cannot then use either of the others as the height:

 

[pre]

.. A

. .

. .

. .

. .

. .

. .

. .

. .

....................

B h C

[/pre]

 

In the case of triangle ABC, above, you can choose to use any side as the base for your formula. Let's choose the bottom one, side BC. You cannot then choose either side AB or AC as the height. Instead, the height of the triangle will now be the length of the line Ah -- a perpendicular line drawn from vertex A, the vertex opposite our base (side BC), to side BC. The perpendicular line Ah will not be as long as either side AB or AC. Consequently, the only true measure of the area of triangle ABC above is:

 

[pre]

BC * Ah / 2

 

BC * AB / 2 and

BC * AC / 2 are both incorrect and will not yield

correct answers

[/pre]

 

Of course, you could choose to use AC or AB as your base. But then you would have to draw a new height line from the opposite vertex to your new base and use that new height in your calculation.

 

The point we've been making above about needing the diagonals is that, in truth, you cannot draw the triangle ABC for the lot in question without knowing more information.

 

We know the lengths of the sides: 159, 157, 133, 110. Suppose we try to figure out the size of the triangle that contains the first two sides. What is the length of the third side?

 

That question can't be answered with the available information because the lot might be shaped like this:

 

[pre]

A B

.............

. .

. .

. .

. .

........

C D

 

or like this:

 

A B

............

. ........

. . D

..

..

.

C

[/pre]

 

I don't have enough control over the placement of the dots, above, to accurately make the four sides the same length in both pictures. But you can clearly see how it can be done. Both lots will have the same four sides, each one the right length. Both lots will have exactly the same perimeter. But they contain vastly different areas.

 

The only way to know which kind of picture we're looking at -- and there are countless variations possible for both types of shape -- is to know the diagonals.

 

In both cases, the diagonal CB will be pretty large. But, the size of the diagonal AD will be large in the first and quite small in the second. Yet both lots have the same perimeter.

 

Note by the way that the second shape is a concave shape precisely because one of its diagonals (CB) lies outside the shape itself.

 

Anyway, I hope this helps.

 

Kind regards,

BG

[Lookin]

I was just over at Caltech's Advanced Applied Mathematics website. They've got a smokin' thread going on Today's Coverboy.

 

Best be careful, guys, we could lose our edge.

[Zippy]

Holy hard-headedness, Batman!!!

 

Please, I beg of you, please stop doing any more math problems. I feel really sorry for your former math teachers.... if they could only see you now.

 

If you would like to learn more about the formula for the area of a triangle, and how to PROPERLY use it, check out the following websites:

 

http://argyll.epsb.ca/jreed/math9/strand3/triangle_area_per.htm

http://www.mathleague.com/help/geometry/area.htm#areaofatriangle

http://www.btinternet.com/~se16/hgb/triangle.htm

http://www.acts.tinet.ie/areaofatriangle_673.html

http://www.mathsisfun.com/area.html

http://www.mste.uiuc.edu/dildine/heron/triarea.html

http://library.thinkquest.org/C0110248/geometry/menareatriangle.htm

http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm#Triangles

[Zippy]

Ok, just call me a glutton for punishment....

 

I am going to attempt one more time to explain how NCM's problem possibly "could", (and I have to emphasize the COULD part)..... possibly "could" work out.

 

Flguy mentions that none of the "guys with the big formulas" has given an answer to the problem yet. Well, there is a very good reason for that. As it has been stated many, many times, we need more information in order to arrive at the exact answer. The problem, stated as such, with the given amount of information is simply impossible to solve. And, when I say "impossible", I am not trying to say that it is difficult or hard to come up with an answer. I am saying that it is truly an unsolvable problem in its current form. Nobody, no matter how smart, or clever, or experienced, or what ever, can give an exact answer to the problem as it stands.

 

Anyway, given the fact that it is unsolvable in its current form, I will still attempt to give it my best guess as to what the other "missing" information "might" be, just so Flguy might see the light.

 

Since most property lots are fairly reasonably shaped, with as many straight parts and right angles as possible, I am going to make some HUGE assumptions here. First, I am going to assume that there are two right angles in the lot. Second, I am going to assume that the numbers that NCM originally gave for this problem were not written down in his message in the order that they sequentially go around the perimeter of the lot. And, thirdly, I am going to assume that the measurements that he gave are only estimates, rounded off answers to the nearest whole foot.

 

Given these assumptions, the following diagram "might" (and once again, I have to emphasize the MIGHT part), be true.

 

.

|..........

|................ 159

|23 ....................

|............................

|______________________

|............157...................|

|....................................|

|....................................|

|110...............................| 110

|....................................|

|....................................|

|....................................|

|....................................|

|______________________|

157

 

 

You'll have to forgive me for the crappy diagram, but I can't for the life of me figure out any other way to draw a decent picture in this message center format.

 

Anyway, for those of you following along at home, as you can see, based on my HUGE assumptions I made above, I have drawn the figure with right angles at the bottom of the figure. Therefore, we can call the base length 157 and the two side lengths 133 (left) and 110 (right). This leaves the 159 length for the top part, which is supposed to be a diagonal line, I just couldn't draw it very well.

 

You can also see that I have drawn another horizontal line across the top of the figure that is supposed to break the figure up into a rectangle (4 sided figure with 4 right angles) and a right triangle at the top. Of course, anyone who knows even the most basic concepts of geometry knows that, using the Pythagorean Theorem, leg squared + leg squared = hypotenuse squared. Well, this is where I am assuming that the measurements for the lengths of the plot of land that NCM gave us are just close approximations, and not exact numbers. Because, if they are exact numbers, then (23)(23) + (157)(157) ≠ (159)(159). But, if you compute the sum of the legs squared (23)(23) + (157)(157) = 25178, and then square root this sum to obtain the hypotenuse of this triangle, you will come up with an answer of 158.68 ft. So, I am assuming that NCM could have rounded this top measurement off to just 159 ft.

 

So, given all of my "major assumptions", then the area of the plot would be equal to the area of the rectangle on the bottom + the area of the triangle on the top. Or,

 

(157)(110) = 17,270

(23)(157)/2 = 1,805.5

17,270 + 1,805.5 = 19,075.5 sq. ft.

 

You may also notice that I am using FLguy's favorite formula of (Base)(Height)/2 to find the area of a triangle. The only difference between his use of this formula and mine is that I used it correctly by actually multiplying the PERPENDICULAR base and height lengths of a triangle together. Not just two random sides of a triangle that are not perpendicular.

 

In any event, all of this is totally a guess with a lot of assumptions thrown in. If NCM actually measures the angles of the lot and comes up with approximately 90º, 90º, 80.9º and 99.1º, then I would be confident of my solution. Or, if he measures the diagonals of the figure and comes up with approximately 205.8 ft and 191.7 ft, I would also have reason to be sure of my answer.

 

But, I will say it ONE MORE TIME, without further information and measurements, no one can accurately say what the answer to the problem is.

 

One thing can be sure, though. And that is, Flguy's attempt to simply take any triangle in the World, with any side lengths and angle measures, and just multiply one of the side lengths times another and divide by 2, will NOT provide the correct answer unless there is a right angle between those two side lengths. In this problem, the right angles could only be consecutive angles, and never be opposite angles since this would force the quadrilateral to be a rectangle. This can't be true because the opposite sides are not equal in length.

 

Lets see if Flguy can wrap his mind around this.......

[Boston Guy]

>I was just over at Caltech's Advanced Applied Mathematics

>website. They've got a smokin' thread going on Today's

>Coverboy.

>

>Best be careful, guys, we could lose our edge.

 

LOL... Ok, you win for funniest post of the day! :-)

 

BG

[Boston Guy]

Well, yes, given the assumptions.

 

I mean, we've known all along that he is probably dealing with something a little shy of half an acre. But the fly in the ointment is the word probably and we can't get rid of it until we have more info to work with.

 

Real world meets theory. You can skip the theory if you want. But then you end up occasionally purchasing something akin to the Brooklyn Bridge.

 

By the way, he could look up a plot diagram on the Internet in many localities. That would answer the question right away.

 

BG

[Lookin]

Thanks. A libido is a terrible thing to waste.