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AWM
05-31-2001, 08:11 PM
I won't try to lie, It's me who needs the brush up. But I goot a get some quick help from you now.


I have a piece of property. And I have to figure out how many acres it is. Problem is there's 5 angles on it, and no even sides. Help me please.


One line is 1020 feet.
another line is 400 feet
another line is 1360 feet.
another line is 950 feet.
and the last line is 160 feet.

I realize that, say,a rectangular lot with 2 even sides is easy enough to figure out.

For example 200 feet by 1000 feet, we would multiply the 2 numbers, and get the product, then I'd divide it by 43560 to get the acres that I need to know. But fu**, throwin un-even lines and and extra angle and I'm stumped.

The formula would be real helpful, and please no flames, we all have our bad subjects. . . .

onalamwar
05-31-2001, 08:42 PM
Hmm.. I don't think there is any one formula for your question. It would help if you had a picture. I think the key is to break up the figure into rectangles and triangles.. If there are 3 right angles in the picture, you could break it into 2 triangles and 1 rectangle. Could you describe how the lines connect? (ie, line 1 connects to line 2 at a 75 degree angle..)

plutarcho
05-31-2001, 08:51 PM
You need to know the angles between the various sides. Then you can break it up into triangles and quadrangles. Then you can use the sine, cosine, tangent formulae to figure the inner sides of the triangles and then the area.

At least, that is where I would begin though there may be an easier way of doing it. But, why do something the easy way when there is surely a more convoluted way to accomplish the same goal.

AWM
05-31-2001, 08:56 PM
http://fs4.dotphoto.com/MemberImages/157366/iBFC3A138-C7DF-4FBA-B8E7-FAA5D708C081.jpg

I appreciate you trying - I can't begin to tell you how long I've sat here, and how many different ways I've tried. They're all wrong so I won't elaborate, but I was on the same track as you.


Here's the link to the "rough" drawing of what it looks like. Thanks alot! I gotta have this done before 10 am tomorrow.

tupacboy
05-31-2001, 09:29 PM
well i don't have much time.. but just looking at ur pic... u have a rectangle that's 1360*60.... a right triangle with 1360 and 360 as it sides and the squareroot of their product (1360*360) as the hypotenuse... and leaves you with a triagle with unknown angles, whose sides are 950, 1020, and the squareroot of (1360*360)

so basically you have
rectangle 1360 by 60
a right triangle with 1360 by 360 with squareroot of 1360*360 being the hypotenuse
and finally a triangle with sides of squareroot 1360 by 360 on one side, 950 on another, and finally 1020 on the last side...

off to do my paper now... i think that's right.. if not sorry
just find the areas of each.. and add them up... :)

tupacboy
05-31-2001, 09:31 PM
hold up... u say it's 160 in ur first message but ur pic shows it as 60... so change all the 360 to 240 if that is the case... and the 60 to 160...

ZrEo0
05-31-2001, 09:39 PM
what level of math is this?

onalamwar
05-31-2001, 09:49 PM
Thanks for the picture!

I think this is how you can get the answer.
Split the figure into 3 parts (I tried to show this in the picture):
http://fs4.dotphoto.com/MemberImages/172785/i61241DAD-350A-46EE-8A89-65CDDFF9A9DB.jpg

On the picture, I'm not sure if its 60ft or 160ft. I assumed 160 ft because that it what you had written in the post.

The first part is a rectangle (1360 by 160). The area of that is simply 1360 * 160 = 217600

The second part is a right triangle. From the picture, the area is 1/2 * a * b. b is 1360, a is 240 (400 for the whole line - 160). So 1/2 * 240 * 1360 = 163200

Before we get to the third part, we need to calculate c. You have to use the Pythagorean theorum. a^2 + b^2 = c^2 (i'm using ^2 to denote squaring). In this case, we have 240^2 + 1360^2 = c^2 Calculating for c, gives approximately 1381.

The third part is a triangle. We know all of the sides (950, 1020, and 1381 which we just calculated). The formula for the area of triangle for which we know all the sides is given by Heron's formula.
Given sides a, b, and c
first calculate s where s = (a+b+c)/2
Then the area is given by Area = SqRt(s(s-a)(s-b)(s-c))

First we calculate s which is (950+1020+1381)/2 = 1675.5

Then we calculate the Area which is:
= SqRt(1675.5 * (1675.5-950) * (1675.5-1020) * (1675.5-1381))
= SqRt(1675.5 * 725.5 * 655.5 * 294.5)
= SqRt(234660420242.4375)
= 484417.6

Summing the three areas (217600 + 163200 + 484418) gives us 865,218 sq. ft.


865,218 sq. ft./ 43560 = 19.86 acres

I wouldn't be surprised if there was a simpler way but I think this is right (as long as the calculations are correct).

Let me know what you think..

Mighty Man
05-31-2001, 09:57 PM
Man, that's a really hard question without knowing 4 of the interior angles, because in theory the sides could bend "inwards" producing a smaller area. If you were to make a pentagon with lengths in similar ration (say a 1/10000th scale model), you'd still be able to "flex" the pentagram, unless you fixed some points.

the closest reference I could find was:
http://www.mathsoft.com/mathcad/library/puzzle/soln30/soln30.html

Best of luck to you...it's got my brain a bit fried.
MM

ZrEo0
05-31-2001, 10:02 PM
there is 5 sides there fore all the angles must add up to 520, i think

Mighty Man
05-31-2001, 10:16 PM
Actually, it's 540 degrees. But without knowing 4 of the interior angles, you cannot know the 5th. You're still making assumptions that don't necessarily hold.

Suppose that the pentagon were redrawn: Instead of having the 950 ft section and the 1020ft section being "external" it would be "internal." The pentagon would be a whole lot smaller with still the same amount of total degrees.

Still confused... =8-(

MM

How about this: Cut the pentagon as drawn into 3 triangles. One line would be drawn from the 1360/60 corner to the 400/1020 corner and the other would be from the 60/950 corner to the 400/1020 corner.

Triangle #1 would have side 1360ft and 400ft and using the pythagoream theorum, the drawn in side should be about 1418 feet.
Triangle #2 would have side 60feet, and we just found that the other side is 1418 feet. We can therefore use the theorum again, and get that the third side should be about 1419 feet (these numbers will vary slightly depending on how you use your sig figs).
Triangle #3 would have side 950feet, 1020ft and the 1419feet that we found earlier (again, the numbers need to be fudged a bit because of rounding).
Once you know the sides of each of the triangles, you should be able to calculate the volume of each of the three triangles and sum them up.

anyone get this or am I missing something obvious?

ZrEo0
05-31-2001, 10:20 PM
Originally posted by Mighty Man
Actually, it's 540 degrees.
my bad, i did the math in my head

Inspect-her Gadget
06-01-2001, 12:40 AM
wow... Got Apexers unite together on a not-so-stupid Off Topic thread. *sniff, sniff* this is a proud moment, guys.

CluelessSi
06-01-2001, 06:54 AM
=)
dunno why u guys split that poor regtangle and tragnle up...... u can just combine them and using the trapazoid area formula to get the area of the simple TRAPAZIOD then u are down to the triangle which u can easily find all the sides to... with the sides u can see if it is a right trangle using ratios... if not then... i think u cut it up to two right trangles and find out the area that way.. i kinda forgot so much of my math skillz =P and with no calculator handy I can't play with the numbers....

DoPeY5007
06-01-2001, 08:24 AM
Originally posted by CluelessSi
=)
and with no calculator handy I can't play with the numbers....



Are you on a PC???...


Start
Programs
Accessories
Calculator

CluelessSi
06-01-2001, 09:00 AM
Originally posted by dopey5007

Originally posted by CluelessSi
=)
and with no calculator handy I can't play with the numbers....



Are you on a PC???...


Start
Programs
Accessories
Calculator

Tried it, it is so much slower then with my TI85 or TI92... Heck I had a hard time looking for squareroot on the windows calculator

AWM
06-01-2001, 09:34 AM
Well friends, we did it. (I) did it according to my boss. He told me he'd drop in at about 10 a.m to go over the sketch and see what I came up with.

Mind you, it was a whole larger project not just that one simple problem I ran into. The rest I tackled and handled like a champ.

Funny thing is after a while last nigth I gave up, and with the age old solution, " I'll do it early tomorrow before he comes in" idea. bahh. Never fu**ing works does it. Well he saved my ass by being late himself. I woke up at about 10 mins. till 10 and I stater going through the math on this thread. Anyway, I finally took what you guys said and what I cam up with was just a nick over 20 acres. So good job. I can't thank you enough for the effort and help. I knew as soon as I was stumped where to turn too.

And lastly, to sum it all up, he didn't want to go over my figures at all. Just asked me the bottom line value for the property and he was satisfied. Love it-=


Take care all!!

AWM
06-01-2001, 09:43 AM
I was just thinking to myself and wanted to make my point a little louder. Special thanks to Mighty Man, Clueless, and onalamwar. You went out of your way to help me. It is appreciated!

tupacboy
06-01-2001, 11:36 AM
gosh i got ignored... sniff sniff... jk... :)