Question 1200168: I know this problem has nothing to do with the chosen topic but I collider find the topic for the mis problem.
Generate a Pythagorean triple using the pair of given numbers and Euclid’s formula.
3 and 8
Found 3 solutions by math_tutor2020, ikleyn, Edwin McCravy:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
For reference
https://en.wikipedia.org/wiki/Pythagorean_triple
Euclid's formula, with respect to the pythagorean triples, refers to a set of 3 equations
- a = m^2 - n^2
- b = 2mn
- c = m^2 + n^2
m & n represent positive integers.
We must have m > n so that m^2-n^2 is positive.
I'll let you confirm those equations satisfy a^2+b^2 = c^2
If m = 8 and n = 3, then,- a = m^2 - n^2 = 8^2 - 3^2 = 64 - 9 = 55
- b = 2mn = 2*8*3 = 48
- c = m^2 + n^2 = 8^2 + 3^2 = 64 + 9 = 73
This makes (55, 48, 73) one possible pythagorean triple out of infinitely many.
a = 55
a^2 = 55^2 = 3025
b = 48
b^2 = 48^2 = 2304
a^2+b^2 = 3025+2304 = 5329
c = 73
c^2 = 73^2 = 5329
This confirms a^2+b^2 = c^2 is true for (a,b,c) = (55,48,73).
The order of a & b doesn't matter. The c must be the longest side.
Answer by ikleyn(52803)  (Show Source):
Answer by Edwin McCravy(20060)  (Show Source):
|
|
|
