STEP 1: find base diagonal $ d $

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:

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

STEP 2: find side $ a $

To find side $ a $ use formula:

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

After substituting $d = 4 \sqrt{ 2 }\, \text{cm}$ we have:

$$ 4 \sqrt{ 2 }\, \text{cm} = \sqrt{ 2 } \cdot a $$ $$ a = \dfrac{ 4 \sqrt{ 2 }\, \text{cm} }{ \sqrt{ 2 } } $$ $$ a = 4\, \text{cm} $$

STEP 3: find slant height $ s $

To find slant height $ s $ use Pythagorean Theorem:

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

After substituting $h = 1\, \text{cm}$ and $a = 4\, \text{cm}$ we have:

$$ \left( 1\, \text{cm} \right)^{2} + \frac{ \left( 4\, \text{cm} \right)^{2} }{ 4 }= s^2 $$ $$ 1\, \text{cm}^2 + \frac{ 16\, \text{cm}^2 }{ 4 }= s^2 $$ $$ 1\, \text{cm}^2 + 4\, \text{cm}^2 = s^2 $$ $$ s^2 = 5\, \text{cm}^2 $$ $$ s = \sqrt{ 5\, \text{cm}^2 } $$$$ s = \sqrt{ 5 }\, \text{cm} $$

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