# math thingy.... i'm not that smart

RatskcorWagoner
Posers and Puzzles 05 Oct '08 19:55
1. 05 Oct '08 19:55
Find the area of a quadrilateral whose vertices have coordinates (-4, -6),

(-3, 5), (4, 7), and (5, -9).
2. 05 Oct '08 20:55
why should I?
3. 05 Oct '08 22:25
Originally posted by afx
why should I?
If you honestly need a reason then don't. I'm just looking for a little help on my math homework. Not that you could help me anyways :]
4. joe shmo
Strange Egg
05 Oct '08 22:352 edits
Originally posted by RatskcorWagoner
If you honestly need a reason then don't. I'm just looking for a little help on my math homework. Not that you could help me anyways :]
are you looking for an analyitical answer?, because I did it graphically

here is what I get for the Area

77 + 49/2 + 11/2 + 192/18 + 225/32 + 77/96 = 125.5

like i said, probably not what your looking for?
5. 06 Oct '08 00:31
Originally posted by joe shmo
are you looking for an analyitical answer?, because I did it graphically

here is what I get for the Area

77 + 49/2 + 11/2 + 192/18 + 225/32 + 77/96 = 125.5

like i said, probably not what your looking for?
I'm not really sure what i am looking for but I know that 125.5 is incorrect.
any other ideas?
6. joe shmo
Strange Egg
06 Oct '08 00:451 edit
Originally posted by RatskcorWagoner
I'm not really sure what i am looking for but I know that 125.5 is incorrect.
any other ideas?
nope...can you point out the flaw in my method? In case your wondering what I did, I plotted the points, connected the lines, and divided the area into a rectangle and several triangles, summing the areas

What is the title of the Math course?

7. 06 Oct '08 00:54
hmm well it is pre calc so i don't know it seems like the logical thing to do..
8. joe shmo
Strange Egg
06 Oct '08 01:05
Originally posted by RatskcorWagoner
hmm well it is pre calc so i don't know it seems like the logical thing to do..
so, what is the area? Was I even close?
9. 06 Oct '08 01:08
If I knew i wouldn't be asking anyone. haha
i will find out the answer tomorrow and will post it.
any other ideas?
10. joe shmo
Strange Egg
06 Oct '08 01:19
Originally posted by RatskcorWagoner
If I knew i wouldn't be asking anyone. haha
i will find out the answer tomorrow and will post it.
any other ideas?
but how do you know it is not 125.5?
11. 06 Oct '08 02:13
I broke it into 2 triangles.

One with sides of sqrt(122), sqrt(53), sqrt(233)...
One with sides of sqrt(257), sqrt(53), sqrt(260).

Using the formula A = 1/4 * sqrt( (A^2 + B^2 + C^2)^2 - 2*(A^4 + B^4 + C^4) ), I get the following..

Triangle 1
Area = 1/4 * Sqrt[(122 + 53 + 233)^2 - 2*(122^2 + 53^2 + 233^2)]
Area = 1/4 * Sqrt[408^2 - 2*(14,884 + 2,809 + 54,289)]
Area = 1/4 * Sqrt(166,464 - 2*71,982)
Area = 1/4 * Sqrt(22,500) = 150 / 4 = 37.5

Triangle 2
Area = 1/4 * Sqrt[(257 + 53 + 260)^2 - 2*(257^2 + 53^2 + 260^2)]
Area = 1/4 * Sqrt[570^2 - 2*(66,049 + 2,809 + 67,600)]
Area = 1/4 * Sqrt(324,900 - 2*136,458)
Area = 1/4 * Sqrt(51,984) = 228 / 4 = 57

Total Area = 37.5 + 57 = 94.5
12. joe shmo
Strange Egg
06 Oct '08 02:36
Originally posted by geepamoogle
I broke it into 2 triangles.

One with sides of sqrt(122), sqrt(53), sqrt(233)...
One with sides of sqrt(257), sqrt(53), sqrt(260).

Using the formula A = 1/4 * sqrt( (A^2 + B^2 + C^2)^2 - 2*(A^4 + B^4 + C^4) ), I get the following..

Triangle 1
Area = 1/4 * Sqrt[(122 + 53 + 233)^2 - 2*(122^2 + 53^2 + 233^2)]
Area = 1/4 * Sqrt[408^2 - 2*(14,884 ...[text shortened]... (324,900 - 2*136,458)
Area = 1/4 * Sqrt(51,984) = 228 / 4 = 57

Total Area = 37.5 + 57 = 94.5
that looks to technical to be false, so you da person geepamoogle.
13. 07 Oct '08 00:58
Pick's Theorem? Although I'll need a piece of graph paper for that.
14. O Artem O
ParTizan
07 Oct '08 02:12