Find the side $ a $ of a rhombus if diagonal $d_1 = 2 \sqrt{ 8 }$ and diagonal $d_2 = 2 \sqrt{ 8 }$.
Answer
$a = 4$
Explanation
To find side $ a $ use formula:
$$ d_1^2 + d_2^2 = 4 \cdot a^2 $$After substituting $ d_1 = 2 \sqrt{ 8 } $ and $ d_2 = 2 \sqrt{ 8 } $ we have:
$$ \left(2 \sqrt{ 8 }\right)^2 + \left(2 \sqrt{ 8 }\right)^2 = 4 \cdot a^2 $$ $$ 32 + 32 = 4 \cdot a^2 $$ $$ 4 \cdot a^2 = 64 $$ $$ a^2 = \frac{ 64 }{ 4 } $$ $$ a^2 = 16 $$ $$ a = \sqrt{ 16 } $$$$ a = 4 $$This page was created using
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