Find the side $ a $ of a rhombus if diagonal $d_1 = \frac{ 97 }{ 8 }$ and diagonal $d_2 = \frac{ 127 }{ 16 }$.

$a = \frac{\sqrt{ 53765 }}{ 32 }$

To find side $ a $ use formula:

$$ d_1^2 + d_2^2 = 4 \cdot a^2 $$

After substituting $ d_1 = \frac{ 97 }{ 8 } $ and $ d_2 = \frac{ 127 }{ 16 } $ we have:

$$ \left(\frac{ 97 }{ 8 }\right)^2 + \left(\frac{ 127 }{ 16 }\right)^2 = 4 \cdot a^2 $$ $$ \frac{ 9409 }{ 64 } + \frac{ 16129 }{ 256 } = 4 \cdot a^2 $$ $$ 4 \cdot a^2 = \frac{ 53765 }{ 256 } $$ $$ a^2 = \frac{ \frac{ 53765 }{ 256 } }{ 4 } $$ $$ a^2 = \frac{ 53765 }{ 1024 } $$ $$ a = \sqrt{ \frac{ 53765 }{ 1024 } } $$$$ a = \frac{\sqrt{ 53765 }}{ 32 } $$

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