Find the lateral area $ L $ of a pyramid if lateral edge $e = 10$ and base diagonal $d = 12$.

$L = 24 \sqrt{ 41 }$

STEP 1: find side $ a $

To find side $ a $ use formula:

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

After substituting $ d = 12 $ we have:

$$ 12 = \sqrt{ 2 } \cdot a $$ $$ a = \dfrac{ 12 }{ \sqrt{ 2 } } $$ $$ a = 6 \sqrt{ 2 } $$

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 = 6 \sqrt{ 2 } $ and $ e = 10 $ we have:

$$ s ^ {\,2} + \frac{ \left(6 \sqrt{ 2 }\right)^2 }{ 4 } = 10^2 $$ $$ s ^ {\,2} + \frac{ 72 }{ 4 } = 10^2 $$ $$ s ^ {\,2} + 18 = 10^2 $$ $$ s ^ {\,2} = 100 - 18 $$ $$ s ^ {\,2} = 82 $$ $$ s = \sqrt{ 82 } $$

STEP 3: find lateral surface $ L $

To find lateral surface $ L $ use formula:

$$ L = 2 \cdot a \cdot s $$

After substituting $ a = 6 \sqrt{ 2 } $ and $ s = \sqrt{ 82 } $ we have:

$$ L = 12 \sqrt{ 2 } \cdot \sqrt{ 82 } $$$$ L = 24 \sqrt{ 41 } $$

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