# perimeter of a triangle

• Jun 13th 2007, 06:50 PM
sanee66
perimeter of a triangle
• Jun 13th 2007, 07:00 PM
Jhevon
Quote:

Originally Posted by sanee66

use the distance formula to find the length of the sides. then just add them together.

recall, the distance formula, that is, in this case, the formula for the length of a line is:

$d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$

do you think you can continue now?
• Jun 13th 2007, 07:27 PM
sanee66
perimeter of a triangle
I come up with 232. Would this be right?
• Jun 13th 2007, 07:32 PM
Jhevon
Quote:

Originally Posted by sanee66
I come up with 232. Would this be right?

hmmm. i got 25.502. i'm not sure how you managed to get that.

triangle DEF has three sides: DE, EF and DF.

we have: $D(-4,6) \mbox { , } E(5,3) \mbox { , } F(3,-2)$

So, the length of DE is:

$DE = \sqrt {(5 - (-4))^2 + (3 - 6)^2} = \sqrt {90}$

the length of EF is:

$EF = \sqrt {(3 - 5)^2 + (-2 - 3)^2} = \sqrt {29}$

the length of DF is:

$DF = \sqrt {(3 - (-4))^2 + (-2 - 6)^2} = \sqrt {113}$

The perimeter is the length of all the sides, therefore

Perimeter $= \sqrt {90} + \sqrt {29} + \sqrt {113} \approx 25.502$
• Jun 13th 2007, 08:07 PM
sanee66
perimeter of a triangle
i was weird and took x1-x2 and y1-y2 instead of the other way around. Thanks for all your help.:D
• Jun 13th 2007, 08:16 PM
ThePerfectHacker
Quote:

Originally Posted by sanee66
i was weird and took x1-x2 and y1-y2 instead of the other way around. Thanks for all your help.:D

It makes no difference.
Since you are squaring
$(y_1-y_2)^2 = (y_2-y_1)^2$