Perimeter of triangle problem

SleepyGurl

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Nov 10, 2006
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Can anyone help me with this hard problem?

A right triangle has a perimeter of 12. What are the lenghts of its 3 sides?

I do know that p=side 1+side2+side3
so 12=Side1+Side2+Side3
so the length of each side would be 4, correct?
 

tkhunny

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Apr 12, 2005
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A "Right" triangle?

It's a perimeter, so Side1+Side2+Side3 = 12. You got that.

Assuming Side3 has the greatest length, there is the Pythagorean Theorem:

(Side1)^2 + (Side2)^2 = (Side3)^2

This is where we have to quit. There is not enough information to do any more. How sure are you that you copied the problem correctly and completely?

Was it a "Right" triangle or an "Equilateral" triangle?
Was it just a "Right" triangle, or an "Isosceles" right triangle?

As stated, there are many, many possibly answers.
 

TchrWill

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Jul 7, 2005
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SleepyGurl said:
Can anyone help me with this hard problem?

A right triangle has a perimeter of 12. What are the lenghts of its 3 sides?

I do know that p=side 1+side2+side3
so 12=Side1+Side2+Side3
so the length of each side would be 4, correct?

The most general formulas for deriving all integer sided right-angled Pythagorean triangles, have been known since the days of Diophantus and the early Greeks. For a right triangle with sides X, Y, and Z, Z being the hypotenuse, the lengths of the three sides of the triangle can be derived as follows: X = k(m^2 - n^2), Y = k(2mn), and Z = k(m^2 + n^2) where k = 1 for primitive triangles (X, Y, and Z having no common factor), m and n are arbitrarily selected integers, one odd, one even, usually called generating numbers, with m greater than n. The symbol ^ means "raised to the power of" such that m^2 means m squared, etc.
Assuming k = 1 and we are seeking integer answers::
m^2 - n^2 + 2mn + m^2 + n^^2 = 12
This reduces to m^2 + mn - 6 = 0
Using the quadratic formula
m = [-n+/-sqrt(n^2 + 24)]/2 = (-n+/-5)/2
For n = 1, m = 2

X = 2^2 - 1^2 = 3
Y = 2(2)1 = 4
Z = 2^2 + 1^2 = 5

3 + 4 + 5 = 12

As tkhunny points out if non-integer answers are acceptable, you have an infinite number of answers.
 

tkhunny

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Very good.

Sleepy, Did the problem state that we were looking for "Integer" or "Whole Number" solutions?
 

SleepyGurl

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Nov 10, 2006
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It doesn't say if it is "Integer" or "Whole Numbers", I totally understand after explaining it, you guys are awesome, thanks so much for your talents and time to help me understand these problems better, THANKS!!!
 
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