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extropalopakettle
No offense, but....

 Posted: Tue Dec 30, 2003 5:17 pm    Post subject: 1 Assume an area of land shaped roughly as follows: code: |\ | \ | \ C | \ | \ | \ | \ | | | | | | | | D| | | | | | | |B | | | | | | | | | | | | | | |_____________| x A y Lengths of the sides are: A = 65.05 B = 279.38 C = 98.45 D = 319.84 Angles x and y are equal (close to but not exactly 90 degrees). What is the area?
The Ktulu
Daedalian Member

 Posted: Tue Dec 30, 2003 6:26 pm    Post subject: 2 Are you sure you put this in the right forum?
extropalopakettle
No offense, but....

 Posted: Tue Dec 30, 2003 6:56 pm    Post subject: 3 Well, I didn't intend it as a puzzle. I know enough information is there. It seems like it should be simple, but I'm not seeing how to solve it.
Dr. Borodog

 Posted: Tue Dec 30, 2003 7:26 pm    Post subject: 4 The area is *approximately* AB + 0.5A(D-B). If x and y are indeed close to 90 degrees, this should probably be sufficient for your purposes. ------------------ You will respect my philosophai.
Laramie
Daedalian Member

 Posted: Tue Dec 30, 2003 7:27 pm    Post subject: 5 You should be able to do it with just the law of sines. Draw a diagonal and apply the law to both triangles. This effectively gives you four equations. Fifth and sixth equations come from setting the sum of the triangle angles to 180. A seventh equation comes from setting the sum of the quadrilateral angles equal to 360. Now draw the other diagonal and repeat the law of sines. This gives you six more equations. What are the unknowns? Three angles of the quad, four angles of the first two triangles, four angles of the second two triangles, and the two lengths of the diagonals (a total of 13 unknowns). Thirteen equations, thirteen unknowns. Once you have the angles, the area would be easy. Of course, I didn't work through it, so I suppose it's possible that the equations aren't all independent.
Quagmire
boring 'n' stuff

 Posted: Tue Dec 30, 2003 7:32 pm    Post subject: 6 But I thought the Law of Sines only worked on right triangles.... ....I could be wrong.
Dr. Borodog

 Posted: Tue Dec 30, 2003 7:34 pm    Post subject: 7 You're wrong. It would be pretty useless if that were true. ------------------ You will respect my philosophai.
mith
Pitbull of Truth

Posted: Tue Dec 30, 2003 8:38 pm    Post subject: 8

 Quote: Fifth and sixth equations come from setting the sum of the triangle angles to 180. A seventh equation comes from setting the sum of the quadrilateral angles equal to 360.

That seventh isn't independent, is it? Maybe I'm picturing something wrong, though.

And even if it was, you can't count that as an equation if you're only counting three of the quad. angles as variables, right? never mind, missed that x and y are equal[/edit]

I suspect it doesn't actually have a unique answer, as quadrilaterals aren't rigid just from side length. But maybe there are just a couple answers, and one could be eliminated by the measures of x and y. same here, though it could still have two answers, one on each side of 90[/edit]

[This message has been edited by mith (edited 12-30-2003 03:45 PM).]
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 8:53 pm    Post subject: 9 What if we pick a point on side D that is the same distance away from x as the length of side B? Then the new quadrilateral would be an isosceles trapezoid, which we can find the area of once we know the new side's length, and we can find the area of a triangle from the three sides as well. The triangle and the trapezoid would each give one equation (law of cosines on the triangle and the difference in the new side and A would depend on x), and there are just two unknowns (x and the new side; the angle of the triangle on side D is congruent to x). I don't have a calculator or anything here, so someone else will have to play with it to see if it actually works.
Laramie
Daedalian Member

 Posted: Tue Dec 30, 2003 8:53 pm    Post subject: 10 Hmmmm, mith. You're right. The sum of the quad angles is not independent. Twelve equations, thirteen unknowns.
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:01 pm    Post subject: 11 Alright, without a calculator so far, so a bunch of letters, but two equations, two unknowns: C^2 = z^2 + (D-B)^2 - 2z(D-B)cos(x) z = A-2Bcos(x) Which should actually have a fairly simple solution for z (well, 2, since/if x is not exactly 90).
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:06 pm    Post subject: 12 0 = D^2 - 2DB + B^2 - BC^2 + (AB-AD)z + Dz^2 [This message has been edited by mith (edited 12-30-2003 04:09 PM).]
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:12 pm    Post subject: 13 ok, so if I've got all this right so far, which I almost certainly don't, z equals: code:AD - AB +/- sqrt(AABB-2AABD+AADD-4DDD+8BDD-4BBD+4BCCD) ------------------------------------------------------ 2D (got tired of the ^s)
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:21 pm    Post subject: 14 cosx = (A-z)/2B Area of quad. = (A+Bcosx)*Bsinx Area of triangle = sqrt(s(s-C)(s-z)(s+B-D)) where s = (C+z+D-B)/2 Someone put all that together.
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:32 pm    Post subject: 15 I think there's something wrong in all that, as I'm not getting very sensible results with excel. Can't be bothered with figuring out what right now, though.
johnny
Cheesy Newbie Abuser

 Posted: Tue Dec 30, 2003 9:32 pm    Post subject: 16 Ok, here we have a trig problem. However, what seems very hard really is just high school math. Forget those blasted sine double angled congugent optical basturds, I think, and make haste to notice 'think' if you split the shape into 3 right angles triangled, in the fashion of two in the recantgle, and the other is left in the top triangular part. As all the sides are known, 0.5base x height should suffice. triangle 1 279.38 x 65.05 x 0.5= 9086.8345 In fact multiply by 2 to get the area of the rectangle, =18173.669 triangle at the top, 40.46 x 65.05 x 0.5=1315.96 total area equals 19489.6 Where am i going wrong then?
mith
Pitbull of Truth

 Posted: Tue Dec 30, 2003 9:36 pm    Post subject: 17 You aren't, that's exactly what Dr. Dawg gave, it just assumes they are exactly 90 degrees, which they aren't.
johnny
Cheesy Newbie Abuser

 Posted: Tue Dec 30, 2003 9:45 pm    Post subject: 18 What i want to know is who invented real life Math problems?
Pablo
Never Draws a Blank

 Posted: Tue Dec 30, 2003 10:40 pm    Post subject: 19 Duh. Like who doesn't want to know that?
extropalopakettle
No offense, but....

Posted: Tue Dec 30, 2003 10:47 pm    Post subject: 20

 Quote: I suspect it doesn't actually have a unique answer, as quadrilaterals aren't rigid just from side length.

It isn't rigid given just the side lengths, but only one way of flexing it makes angles x and y equal. Given x and y are equal, I'm sure there's a unique answer. I could write a program. Given lengths A, B and D and a value for x=y, there's a unique value for length C ... so just find x=y (by searching) that gives the given value of C. Once angles x and y are known, the rest is easy. But that's cheating.
extropalopakettle
No offense, but....

 Posted: Tue Dec 30, 2003 11:43 pm    Post subject: 21 Oh, I geuss I should point out that I linked to this thread over here: http://www.greylabyrinth.com/Forums/Forum5/HTML/005112.html?3 where there is seperate discussion. I'm working through kevinatilusa's approach at the moment.
Dr. Borodog

 Posted: Wed Dec 31, 2003 1:55 am    Post subject: 22 Are x & y acute or obtuse? ------------------ You will respect my philosophai.
extropalopakettle
No offense, but....

 Posted: Wed Dec 31, 2003 1:58 am    Post subject: 23 Greater than 90 degrees. Hmmm ... can there be a solution with them less than 90?
zeek
Daedalian Member

 Posted: Wed Dec 31, 2003 3:12 am    Post subject: 24 Is this problem done yet? I just did it on a spreadsheet and got an area of 21307. I think it's in the ballpark, since A*(B+D)/2 ~ 19435. (I'm not supposed to do math this week since I'm on vacation.) Does anyone else get 21307?
zeek
Daedalian Member

 Posted: Wed Dec 31, 2003 3:15 am    Post subject: 25 Oops. Just fixed a problem. Make that 23155 that I got. (I told you I should be on vacation!)
mith
Pitbull of Truth

Posted: Wed Dec 31, 2003 11:39 am    Post subject: 26

 Quote: Greater than 90 degrees. Hmmm ... can there be a solution with them less than 90?

Yeah, there *could* be, I think. When I was first speculating on multiple answers I hadn't read that x=y, but even with that, I suspect there may be answers on each side of 90. Like how SSA isn't a theorem.

And waaah, kevin's method is the same as mine. Just 'cause I use letters all over the place everyone gets scared...
mith
Pitbull of Truth

 Posted: Wed Dec 31, 2003 11:58 am    Post subject: 27 Yep, after sleeping, and typing everything back into excel, I get 23155.80611, which matches what zeek got. Did you use the same method, zeek?
zeek
Daedalian Member

 Posted: Wed Dec 31, 2003 12:45 pm    Post subject: 28 Yes, mith, I pretty much came up with the same approach that you and kevin had, with some insignificant variations. In particular, I did this: Where line B meets line C, call it point z. Where line C meets line D, call it point w. Draw a line through w parallel to A until it meets line B extended at point v. Drop perpendicular lines from w and v that meet line A extended at points p and q, respectively. Call line segment wv E. Call line segment zv F. Call angles
Dr. Borodog

 Posted: Wed Dec 31, 2003 4:54 pm    Post subject: 29 And what do you get for x and y? ------------------ You will respect my philosophai.
CFBH
Icarian Member

 Posted: Wed Dec 31, 2003 5:13 pm    Post subject: 30 I get x = y = 92.36407577 deg There is no solution for an acute angle.
zeek
Daedalian Member

 Posted: Wed Dec 31, 2003 5:48 pm    Post subject: 31 For angles x&y I got what CFBH said. Actually I bet there is a solution where x and y are acute, except that I think then sides B and D would be crossing each other, forming an X between sides A and C in a figure-8 type thingy. In fact, that quadratic equation produced two values for cos(t), one of which gives a value of 74.96 for x & y. I didn't check, but I bet that is the acute solution with the weird shape.
CFBH
Icarian Member

 Posted: Wed Dec 31, 2003 9:24 pm    Post subject: 32 Ha! - I was waiting for that. Of course mathematically you are correct - if you allow B and D to cross you get 2 triangles touching at the intersection, and 2 separate areas. But this is a real world problem - it seems to me that extropalopakettle's orignal question related to a plot of land, which by definition cannot have its own boundaries crossing.
mith
Pitbull of Truth

 Posted: Wed Dec 31, 2003 9:30 pm    Post subject: 33 Yes, but in theory it could have two solutions, anyway. No point doing a problem like this if you can't find a general solution. [This message has been edited by mith (edited 12-31-2003 04:31 PM).]
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