Sides of a triangle.

Surfer

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Mar 2, 2006
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I figured this has been asked several times, but I didn't find one similar to it.

Perimeter of a triangle is 210 meters. The longest side is 50 meters longer than the shortest side. The remaining side is twice as long as the shortest side. Find the length of each side.

This is what I have.

A + B +C = 210
A = 50 + B
C = 2B

But I don't know which one to plug in etc.
 
Surfer said:
I figured this has been asked several times, but I didn't find one similar to it.

Perimeter of a triangle is 210 meters. The longest side is 50 meters longer than the shortest side. The remaining side is twice as long as the shortest side. Find the length of each side.

This is what I have.

A + B +C = 210
A = 50 + B
C = 2B

But I don't know which one to plug in etc.

Your setup is correct, but you should define things more clearly.

i.e.
Let A be the longest side
Let B be the shortest side (according to your setup)
Let C be the remaining side

A + B + C = 210
A = B + 50
C = 2*B

Notice that A and C are described as "functions" of B and only B.

So A + B + C = (B+50) + B + (2*B) = 210.

Hope that helps,
Daon
 
Surfer said:
Perimeter of a triangle is 210 meters. The longest side is 50 meters longer than the shortest side. The remaining side is twice as long as the shortest side. Find the length of each side.
This is what I have.
A + B +C = 210
A = 50 + B
C = 2B
But I don't know which one to plug in etc.
Well, what you have is ok, but you only need 1 variable:
let x = short side
then longest = x + 50
and remaining = 2x

SO: x + x + 50 + 2x = 210
ok?
 
Thanks for the help guys. I figured it out, now I'm moving on to an icosceles triangle, this will be helpful.

Thanks again. :)
 
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