Enter three numbers to check for Pythagorean triples, or input two sides to calculate the third side.
Pythagorean triples are three positive integers \(a\), \(b\), and \(c\) that satisfy the Pythagorean theorem: \( c^2 = a^2 + b^2\) Here, \(c\) is the hypotenuse (the longest side), and \(a\) and \(b\) are the two legs of a right triangle. If this relationship holds true, the three numbers are considered a Pythagorean triple.
To calculate the hypotenuse (\(c\)) when both legs (\(a\) and \(b\)) are known: \( c = \sqrt{a^2 + b^2} \)
To calculate one leg (\(b\)) when the other leg (\(a\)) and the hypotenuse (\(c\)) are known: \( b = \sqrt{c^2 - a^2} \)
Verify:
\( 3^2 + 4^2 = 9 + 16 = 25 \)
\( 5^2 = 25 \)
\( 3^2 + 4^2 = 5^2 \)
Yes, \(3\), \(4\), and \(5\) form a Pythagorean triple.
Verify:
\( 30^2 + 40^2 = 900 + 1600 = 2500 \)
\( 45^2 = 2025 \)
\( 30^2 + 40^2 \neq 45^2 \)
No, \(30\), \(40\), and \(45\) do not form a Pythagorean triple.
Solution:
Hypotenuse (\(c\)):
\(c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)
The hypotenuse is \(10\), forming a Pythagorean triple \(6, 8, 10\).
Alternate case (other leg, not the hypotenuse):
\(c = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 5.29\)
It is not an integer and cannot be a Pythagorean trigon.