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

Find the area of a quadrilateral whose vertices have coordinates (-4, -6),

(-3, 5), (4, 7), and (5, -9).

why should I?

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 :]

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?

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?

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?

here is a link with some information:http://mathworld.wolfram.com/Quadrilateral.html

hmm well it is pre calc so i don't know it seems like the logical thing to do..

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?

If I knew i wouldn't be asking anyone. haha
i will find out the answer tomorrow and will post it.
any other ideas?

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?

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

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.

Pick's Theorem? Although I'll need a piece of graph paper for that.

so whats the answer?

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?

Are we assuming the first # to be the X axis, and second # to be Y?
I got the 77, but I got 8+5.5+7+12+77 ,minus two small areas, which comes out to 109 and change.