05 Oct '08 19:55

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

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

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

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podunk, PA05 Oct '08 22:352 edits

are you looking for an analyitical answer?, because I did it graphically*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 :]**

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?- Joined
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06 Oct '08 00:31

I'm not really sure what i am looking for but I know that 125.5 is incorrect.*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?

any other ideas?- Joined
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podunk, PA06 Oct '08 00:451 edit

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*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?

What is the title of the Math course?

here is a link with some information:http://mathworld.wolfram.com/Quadrilateral.html- Joined
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06 Oct '08 02:13I 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- Joined
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podunk, PA06 Oct '08 02:36

that looks to technical to be false, so you da person geepamoogle.*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- Joined
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slatington, pa, usa07 Oct '08 03:42

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