Find the lateral area $ L $ of a pyramid if height $h = 25$ and base area $B = 11$.

$L = 9 \sqrt{ 341 }$

STEP 1: find side $ a $

To find side $ a $ use formula:

After substituting $ B = 11 $ we have:

$$ B = a^2 $$ $$ a^2 = 11 $$ $$ a = \sqrt{ 11 } $$

STEP 2: find slant height $ s $

To find slant height $ s $ use Pythagorean Theorem:

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

After substituting $ h = 25 $ and $ a = \sqrt{ 11 } $ we have:

$$ 25^2 + \frac{ \left( \sqrt{ 11 } \right)^{ 2 } }{ 4 }= s^2 $$ $$ 625 + \frac{ 11 }{ 4 }= s^2 $$ $$ 625 + \frac{ 11 }{ 4 } = s^2 $$ $$ s^2 = \frac{ 2511 }{ 4 } $$ $$ s = \sqrt{ \frac{ 2511 }{ 4 } } $$$$ s = \frac{ 9 \sqrt{ 31}}{ 2 } $$

STEP 3: find lateral surface $ L $

To find lateral surface $ L $ use formula:

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

After substituting $ a = \sqrt{ 11 } $ and $ s = \frac{ 9 \sqrt{ 31}}{ 2 } $ we have:

$$ L = 2 \sqrt{ 11 } \cdot \frac{ 9 \sqrt{ 31}}{ 2 } $$$$ L = 9 \sqrt{ 341 } $$

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