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

$s = \frac{\sqrt{ 394 }}{ 2 }$

STEP 1: find side $ a $

To find side $ a $ use formula:

After substituting $ B = 6 $ we have:

$$ B = a^2 $$ $$ a^2 = 6 $$ $$ a = \sqrt{ 6 } $$

STEP 2: find slant height $ s $

To find slant height $ s $ use Pythagorean Theorem:

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

After substituting $ a = \sqrt{ 6 } $ and $ e = 10 $ we have:

$$ s ^ {\,2} + \frac{ \left( \sqrt{ 6 } \right)^{ 2 } }{ 4 } = 10^2 $$ $$ s ^ {\,2} + \frac{ 6 }{ 4 } = 10^2 $$ $$ s ^ {\,2} + \frac{ 3 }{ 2 } = 10^2 $$ $$ s ^ {\,2} = 100 - \frac{ 3 }{ 2 } $$ $$ s ^ {\,2} = \frac{ 197 }{ 2 } $$ $$ s = \sqrt{ \frac{ 197 }{ 2 } } $$$$ s = \frac{\sqrt{ 394 }}{ 2 } $$

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