real world math problem

[extropalopakettle]

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]

Are you sure you put this in the right forum?

[extropalopakettle]

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]

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]

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]

But I thought the Law of Sines only worked on right triangles....


....I could be wrong.

[Dr. Borodog]

You're wrong. It would be pretty useless if that were true.



------------------
You will respect my philosophai.

[mith]

[mith]

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]

Hmmmm, mith. You're right. The sum of the quad angles is not independent. Twelve equations, thirteen unknowns.

[mith]

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]

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]

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]

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]

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]

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]

You aren't, that's exactly what Dr. Dawg gave, it just assumes they are exactly 90 degrees, which they aren't.

[johnny]

What i want to know is who invented real life Math problems?

[Pablo]

Duh. Like who doesn't want to know that?

[extropalopakettle]

[extropalopakettle]

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]

Are x & y acute or obtuse?


------------------
You will respect my philosophai.

[extropalopakettle]

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

[zeek]

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]

Oops. Just fixed a problem. Make that 23155 that I got.
(I told you I should be on vacation!)

[mith]

[mith]

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]

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 <wxp and <vyq 't'(for 'theta'?) and call cos(t) T.

Since E is parallel to A, then F is just D-B.
Also, angle <wvz = t.
E is A + 2DT.
By the law of cosines,
EE + FF - 2EFT = CC.
This makes a messy quadratic in T, and you solve for T.
Compute the final area by computing the area of rectangle wvqp, then subtract the two side triangles wxp and vyq, then subtract top triangle wvz.

[Dr. Borodog]

And what do you get for x and y?


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You will respect my philosophai.

[CFBH]

I get x = y = 92.36407577 deg
There is no solution for an acute angle.

[zeek]

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]

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]

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).]