welcome to my nightmare

purplemonkey

New member
Joined
Sep 20, 2005
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6
So, I am sure that this will be an easy question for ya'll tonight, I havent had a math class in a million years and I am trying to complete some much needed extra credit work, but I have gotten a little frustrated.

The question is.... A triangle has a perimeter of 161 miles, each of the two smaller sides of the triangle are two-thirds the lenght of the longest side. Find the lenght of each side.....
So, heres what I know....
I know that the perimeter is the lenght of all the sides togeather....
a + b + c = p
I can tell that two sides are the same lenght here so lets say a =b
and those are 2/3 of the side c....so should my equation look like

(c-1/3) + (c-1/3) + c =161 ??
and if that is right how in the world do i solve that? can i say 3c-2/3 =161? i am sure that isnt right....i am really going a bit crazy.... my math book doesnt have a geometry chapter really, so i am lost.
 
They aren't "c units long, less 1/3 of a unit"; they are "two-thirds of the length of c". "c - 1/3" says the former; "(2/3)c" says the latter.

You're on the right track. Now make the above correction, and see where that takes you.

Eliz.
 
A triangle has a perimeter of 161 miles, each of the two smaller sides of the triangle are two-thirds the lenght of the longest side. Find the lenght of each side

If the longest side is x, then 2x/3 + 2x/3 + x = 161 or 7x/3 = 161.

I suspect you can take it from here.
 
ok....lets see

i think my biggest weekness would be dealing with fractions....

Ok...

2/3c + 2/3c+c=161

2 1/3c=161

c=69 161-69=92 92/2=46

so the sides are 69, 46 and 46?

i might ask you for help with a uniform motion problem next... :)
if i am rigth and didnt screw that up thanks a ton
 
To check the answer to any "solving" problem, plug the solution back into the original problem. If it "works", the solution is probably valid.

Eliz.
 
It is simpler to leave it as an improper fraction.
2/3c + 2/3c+c=161
(2 +2 +3)c/3=
7c/3 = 161
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Try it that way next time.
 
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