Pure Pythagoras

Let $E$ be the foot of the pyramid's height $SE$ , $a$ the length of the side $AB$ and $b$ the length of the side $BC$ of the pyramid's rectangular basis. Drop four lines from $E$ so that each of them is perpendicular to the lines that contain each of $ABCD$ 's sides, and let $p$ be the length of the line perpendicular to line $AB$ and $q$ the length of the line perpendicular to line $BC$ . In case $E$ is inside $ABCD$ , by Pythagoras' Theorem, we have:

• $AE^2 = p^2 + q^2$

• $BE^2 = (a - p)^2 + q^2$

• $CE^2 = (a - p)^2 + (b - q)^2$

• $DE^2 = p^2 + (b - q)^2$

Sum up first and third equation, and then second and fourth to get:

$AE^2 + CE^2 = BE^2 + DE^2$

In case $E$ is outside $ABCD$ , in equations above we would use $a + p$ or $b + q$ depending on where the point $E$ is, but that doesn't affect the result.
Again, by Pythagoras' Theorem:

• $SA^2 = AE^2 + SE^2$

• $SB^2 = BE^2 + SE^2$

• $SC^2 = CE^2 + SE^2$

• $SD^2 = DE^2 + SE^2$

Now:

$SA^2 + SC^2 = AE^2 + CE^2 + 2SE^2 = BE^2 + DE^2 + 2SE^2 = SB^2 + SD^2$

So:

$\boxed{SA^2 + SC^2 = SB^2 + SD^2}$

From this we can easily calculate the length of $SD$ :

$SD = \sqrt{SA^2 - SB^2 + SC^2} = \sqrt{5 - 8 + 4} = \sqrt{1} = \boxed{1}$