A simple problem that's dirving me crazy (Perimeter and length of each side)

[phoenix20_06]

Homework Statement

A city lot has the shape of a right triangle whose hypotehenuse is 7 ft longer that one of the other sides. Perimiter of the lot is 392 ft how long is each side.

Homework Equations

P = a + b +c = 392

hypothenuse is c^2 = a^2 + b^2

so if c is 7ft longer then one of the sides (let's say b)

then b +7 = c or b = c-7

a+ c-7 +c = 392
a +2c = 399
a = 399 -2c

To find c I use

Pythogorian formula but replace a and b with their equivalent of c

(399-2c)^2 + (c-7)^2 = c^2

Is it the right way to resolve the problem? if it is, why after 4-5 attempts I keep getting different values of c or sometimes no c at all :)

If it's the right way to go

is the (399-2c)^2 + (c-7)^2 = c^2 implies...

-4c^2 + 1596c + 159201 +c^2 - 14c + 49 = c^2 ? Just to make sure I'm not missing a minus.

Thank you:)

The Attempt at a Solution

 

[cristo]

Your method looks right.

-4c^2 + 1596c + 159201 +c^2 - 14c + 49 = c^2 ? Just to make sure I'm not missing a minus.

This should read +4c²+...

 

[phoenix20_06]

okay this what I get then...
4c^2 + 1582c + 159250 = 0
or
2c^2 + 791c + 79625 = 0 if I divide everything by 2

c = -b +/- sqrt of(b^2 -4ac) and everything divided by 2a

however b^2 - 4ac is negative and I can't do a sqrt of a negative number.
What went wrong?

791^2 - 4(2)(79625) = 625681 - 637000 = -11319

 

[cristo]

-4c^2 + 1596c + 159201 +c^2 - 14c + 49 = c^2

There's another mistake in here: this should read 4c²-1596c+...

 

[phoenix20_06]

What the deuce! It's wichcraft not mathematics! I just solved it
the sides are c = 175, b = 168 and a = 49

49^2 + 168^2 = 175^2

Thank you Cristo, I think your presence in my posts make me think twice harder :)

Those minuses messed up calculations in the first place :)

 

[cristo]

You're welcome!