# More Algebra

• Aug 20th 2008, 02:32 PM
mathgeek777
More Algebra
Quote:

A city lot has the shape of a right triangle whose hypotenuse is 7 ft longer than one of the other sides. The perimeter of the lot is 392 ft. How long is each side of the lot?
Let x be one side, ? be another, and x + 7 be the hypotenuse. You find that ? = $\sqrt{14x+49}$. Plug all 3 into the perimeter formula with 392 as the perimeter.

After some math, I got $14x + 49 = 4x^2 - 1540x + 148225$. Subtract 14x + 49 and I get $0=4x^2-1554x+148176$

At this point I'm lost because the numbers I'm getting after plugging it into the quadratic equation are anywhere between 80-90% of the perimeter, which doesn't make sense. If anyone could help, that'd be great.
• Aug 20th 2008, 03:20 PM
galactus
The length is $x^{2}+y^{2}=(x+7)^{2}$

The perimeter is $x+y+x+7=2x+y+7=392$

$y=385-2x$

Sub into the area formula:

$x^{2}+(385-2x)^{2}=(x+7)^{2}$. What you have.

Solve for x and we get x=168 and x=220.5

The 168 is the good answer. Because 385-2(220.5) would result in a negative side length.

So, we have side lengths of 168 and 49 and 175
• Aug 20th 2008, 03:32 PM
Chop Suey
Your work is fine. You could have used the quadratic formula to find the roots.

$4x^2 - 1554x + 148176 = 0$

$a = 4, b = -1554, c = 148176$

$x = \frac{1554\pm\sqrt{(-1554)^2 - 4(4)(148176)}}{2(4)}$

$x = \frac{1554\pm\sqrt{44,100}}{8}$

$x = \frac{1554\pm210}{8}$

$x = 220.5\ or\ x = 168$

Only of the the roots gives the correct answer.

EDIT: aaah, galactus you beat me. :p
• Aug 20th 2008, 03:39 PM
mathgeek777
It's cool, I'll give you both thanks.

I don't know where I was getting the ridiculous numbers from. Maybe I did something wrong in my work. Not sure.