Find the perimeter $ P $ of a rhombus if diagonal $d_1 = 17$ and diagonal $d_2 = 8$.
Answer
$P = 2 \sqrt{ 353 }$
Explanation
STEP 1: find side $ a $
To find side $ a $ use formula:
$$ d_1^2 + d_2^2 = 4 \cdot a^2 $$After substituting $ d_1 = 17 $ and $ d_2 = 8 $ we have:
$$ 17^2 + 8^2 = 4 \cdot a^2 $$ $$ 289 + 64 = 4 \cdot a^2 $$ $$ 4 \cdot a^2 = 353 $$ $$ a^2 = \frac{ 353 }{ 4 } $$ $$ a^2 = \frac{ 353 }{ 4 } $$ $$ a = \sqrt{ \frac{ 353 }{ 4 } } $$$$ a = \frac{\sqrt{ 353 }}{ 2 } $$STEP 2: find perimeter $ P $
To find perimeter $ P $ use formula:
$$ P = 4 \cdot a $$After substituting $ a = \frac{\sqrt{ 353 }}{ 2 } $ we have:
$$ P = 4 \cdot \frac{\sqrt{ 353 }}{ 2 } $$ $$ P = 2 \sqrt{ 353 } $$This page was created using
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