To find base diagonal $ d $ use Pythagorean Theorem:

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

After substituting $h = 1\, \text{cm}$ and $e = 3\, \text{cm}$ we have:

$$ \left( 1\, \text{cm} \right)^{2} + \frac{ d^2 }{ 4 } = \left( 3\, \text{cm} \right)^{2} $$ $$ \frac{ d^2 }{ 4 } = \left( 3\, \text{cm} \right)^{2} - \left( 1\, \text{cm} \right)^{2} $$ $$ \frac{ d^2 }{ 4 } = 9\, \text{cm}^2 - 1\, \text{cm}^2 $$ $$ d^2 = 8\, \text{cm}^2 \cdot 4 $$ $$ d^2 = 32\, \text{cm}^2 $$ $$ d = \sqrt{ 32\, \text{cm}^2 } $$$$ d = 4 \sqrt{ 2 }\, \text{cm} $$

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