Find the height $ h $ of a pyramid if side $a = 16$ and lateral edge $e = 2 \sqrt{ 41 }$.

$h = 6$

STEP 1: find base diagonal $ d $

To find base diagonal $ d $ use formula:

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

After substituting $ a = 16 $ we have:

$$ d = \sqrt{ 2 } \cdot 16 $$ $$ d = 16 \sqrt{ 2 } $$

STEP 2: find height $ h $

To find height $ h $ use Pythagorean Theorem:

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

After substituting $ d = 16 \sqrt{ 2 } $ and $ e = 2 \sqrt{ 41 } $ we have:

$$ h ^ {\,2} + \frac{ \left(16 \sqrt{ 2 }\right)^2 }{ 4 } = \left(2 \sqrt{ 41 }\right)^2 $$ $$ h ^ {\,2} + \frac{ 512 }{ 4 } = \left(2 \sqrt{ 41 }\right)^2 $$ $$ h ^ {\,2} + 128 = \left(2 \sqrt{ 41 }\right)^2 $$ $$ h ^ {\,2} = 164 - 128 $$ $$ h ^ {\,2} = 36 $$ $$ h = \sqrt{ 36 } $$$$ h = 6 $$

Pyramid Calculator