Find the slant height $ s $ of a pyramid if height $h = 6$ and lateral edge $e = 10$.

$s = 2 \sqrt{ 17 }$

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 = 6 $ and $ e = 10 $ we have:

$$ 6^2 + \frac{ d^2 }{ 4 } = 10^2 $$ $$ \frac{ d^2 }{ 4 } = 10^2 - 6^2 $$ $$ \frac{ d^2 }{ 4 } = 100 - 36 $$ $$ d^2 = 64 \cdot 4 $$ $$ d^2 = 256 $$ $$ d = \sqrt{ 256 } $$$$ d = 16 $$

STEP 2: find side $ a $

To find side $ a $ use formula:

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

After substituting $ d = 16 $ we have:

$$ 16 = \sqrt{ 2 } \cdot a $$ $$ a = \dfrac{ 16 }{ \sqrt{ 2 } } $$ $$ a = 8 \sqrt{ 2 } $$

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 = 6 $ and $ a = 8 \sqrt{ 2 } $ we have:

$$ 6^2 + \frac{ \left(8 \sqrt{ 2 }\right)^2 }{ 4 }= s^2 $$ $$ 36 + \frac{ 128 }{ 4 }= s^2 $$ $$ 36 + 32 = s^2 $$ $$ s^2 = 68 $$ $$ s = \sqrt{ 68 } $$$$ s = 2 \sqrt{ 17 } $$

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