Pythagorean theorem

c = √(a²+b²) · leg · 3D diagonal

Side aSide b

Hypotenuse c

In a right triangle: c² = a² + b² (c — the hypotenuse, a, b — the legs).

The Pythagorean theorem relates the sides of a right triangle:

$c^2=a^2+b^2$

where $c$ is the hypotenuse (the longest side) and $a, b$ are the legs. Hence $c=\sqrt{a^2+b^2}$ , and a missing leg is found as $b=\sqrt{c^2-a^2}$ .

In space the theorem is applied twice, giving the space diagonal of a cuboid:

$d=\sqrt{a^2+b^2+c^2}$

The calculator handles all three cases. The classic Pythagorean triple is 3-4-5 (since 3²+4²=5²).

Examples

Legs a=3, b=4

→ c = √(9+16) = 5

Cuboid 2×3×6

→ diagonal = √(4+9+36) = 7

Which side is the hypotenuse?+

The longest one — the side opposite the right angle. It is always c in the formula c² = a² + b². The legs (a, b) form the right angle.

How do I compute the space diagonal of a cuboid?+

By Pythagoras twice: first the base diagonal √(a²+b²), then the diagonal of the solid √((√(a²+b²))² + c²) = √(a²+b²+c²).

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