Find the height $ h $ of a pyramid if base diagonal $d = 5$ and total surface $A = 125$.

$h = 5 \sqrt{ 10 }$

STEP 1: find side $ a $

To find side $ a $ use formula:

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

After substituting $ d = 5 $ we have:

$$ 5 = \sqrt{ 2 } \cdot a $$ $$ a = \dfrac{ 5 }{ \sqrt{ 2 } } $$ $$ a = \frac{ 5 \sqrt{ 2}}{ 2 } $$

STEP 2: find base area $ B $

To find base area $ B $ use formula:

$$ B = a^2 $$

After substituting $ a = \frac{ 5 \sqrt{ 2}}{ 2 } $ we have:

$$ B = \left( \frac{ 5 \sqrt{ 2}}{ 2 } \right)^2 $$ $$ B = \frac{ 25 }{ 2 } $$

STEP 3: find lateral surface $ L $

To find lateral surface $ L $ use formula:

$$ A = B + L $$

After substituting $ A = 125 $ and $ B = \frac{ 25 }{ 2 } $ we have:

$$ 125 = \frac{ 25 }{ 2 } + L $$ $$ L = 125 - \frac{ 25 }{ 2 } $$ $$ L = \frac{ 225 }{ 2 } $$

STEP 4: find slant height $ s $

To find slant height $ s $ use formula:

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

After substituting $ L = \frac{ 225 }{ 2 } $ and $ a = \frac{ 5 \sqrt{ 2}}{ 2 } $ we have:

$$ \frac{ 225 }{ 2 } = 2 \cdot \frac{ 5 \sqrt{ 2}}{ 2 } \cdot s $$$$ \frac{ 225 }{ 2 } = 5 \sqrt{ 2 } \cdot s $$$$ s = \dfrac{ \frac{ 225 }{ 2 } }{ 5 \sqrt{ 2 } } $$$$ s = \frac{ 45 \sqrt{ 2}}{ 4 } $$

STEP 5: find height $ h $

To find height $ h $ use Pythagorean Theorem:

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

After substituting $ a = \frac{ 5 \sqrt{ 2}}{ 2 } $ and $ s = \frac{ 45 \sqrt{ 2}}{ 4 } $ we have:

$$ h ^ {\,2} + \frac{ \left( \frac{ 5 \sqrt{ 2}}{ 2 } \right)^2 }{ 4 } = \left( \frac{ 45 \sqrt{ 2}}{ 4 } \right)^2 $$ $$ h ^ {\,2} + \frac{ \frac{ 25 }{ 2 } }{ 4 } = \left( \frac{ 45 \sqrt{ 2}}{ 4 } \right)^2 $$ $$ h ^ {\,2} + \frac{ 25 }{ 8 } = \left( \frac{ 45 \sqrt{ 2}}{ 4 } \right)^2 $$ $$ h ^ {\,2} = \frac{ 2025 }{ 8 } - \frac{ 25 }{ 8 } $$ $$ h ^ {\,2} = 250 $$ $$ h = \sqrt{ 250 } $$$$ h = 5 \sqrt{ 10 } $$

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