STEP 1: find base diagonal $ d $

To find base diagonal $ d $ use formula:

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

After substituting $a = 15\, \text{cm}$ we have:

$$ d = \sqrt{ 2 } \cdot 15\, \text{cm} $$ $$ d = 15 \sqrt{ 2 }\, \text{cm} $$

STEP 2: find height $ h $

To find height $ h $ use formula:

$$ V = \dfrac{ a ^{ 2 } \cdot h }{ 3 } $$

After substituting $V = 900\, \text{cm}$ and $a = 15\, \text{cm}$ we have:

$$ 900\, \text{cm} = \dfrac{ 15\, \text{cm} ^{ 2 } \cdot h }{ 3 } $$$$ 900\, \text{cm} \cdot 3 = 15\, \text{cm} ^{ 2 } \cdot h $$$$ 2700\, \text{cm} = 225\, \text{cm}^2 \cdot h $$$$ h = \dfrac{ 2700\, \text{cm} }{ 225\, \text{cm}^2 } $$$$ h = 12\, \text{cm}^-1 $$

STEP 3: find lateral edge $ e $

To find lateral edge $ e $ use Pythagorean Theorem:

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

After substituting $h = 12\, \text{cm}^-1$ and $d = 15 \sqrt{ 2 }\, \text{cm}$ we have:

$$ \left( 12\, \text{cm}^-1 \right)^{2} + \frac{ \left( 15 \sqrt{ 2 }\, \text{cm} \right)^{2} }{ 4 }= e^2 $$ $$ 144\, \text{cm}^-2 + \frac{ 450\, \text{cm}^2 }{ 4 }= e^2 $$ $$ 144\, \text{cm}^-2 + \frac{ 225 }{ 2 }\, \text{cm}^2 = e^2 $$ $$ e^2 = \frac{ 513 }{ 2 }\, \text{cm}^-2 $$ $$ e = \sqrt{ \frac{ 513 }{ 2 }\, \text{cm}^-2 } $$$$ e = \frac{ 3 \sqrt{ 114}}{ 2 }\, \text{cm}^-1 $$

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