Find the slant height $ s $ of a pyramid if height $h = 1$ and lateral edge $e = 3$.

$s = \sqrt{ 5 }$

STEP 1: find base diagonal $ d $

To find base diagonal $ d $ use Pythagorean Theorem:

$$ h^2 + \frac{ d^2 }{ 4 } = e^2 $$

After substituting $ h = 1 $ and $ e = 3 $ we have:

$$ 1^2 + \frac{ d^2 }{ 4 } = 3^2 $$ $$ \frac{ d^2 }{ 4 } = 3^2 - 1^2 $$ $$ \frac{ d^2 }{ 4 } = 9 - 1 $$ $$ d^2 = 8 \cdot 4 $$ $$ d^2 = 32 $$ $$ d = \sqrt{ 32 } $$$$ d = 4 \sqrt{ 2 } $$

STEP 2: find side $ a $

To find side $ a $ use formula:

$$ d = \sqrt{ 2 } \cdot a $$

After substituting $ d = 4 \sqrt{ 2 } $ we have:

$$ 4 \sqrt{ 2 } = \sqrt{ 2 } \cdot a $$ $$ a = \dfrac{ 4 \sqrt{ 2 } }{ \sqrt{ 2 } } $$ $$ a = 4 $$

STEP 3: find slant height $ s $

To find slant height $ s $ use Pythagorean Theorem:

$$ h^2 + \frac{ a^2 }{ 4 } = s^2 $$

After substituting $ h = 1 $ and $ a = 4 $ we have:

$$ 1^2 + \frac{ 4^2 }{ 4 }= s^2 $$ $$ 1 + \frac{ 16 }{ 4 }= s^2 $$ $$ 1 + 4 = s^2 $$ $$ s^2 = 5 $$ $$ s = \sqrt{ 5 } $$

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