Find the perimeter $ P $ of a rhombus if diagonal $d_2 = 25$ and angle $\alpha = 60^o$.

$P = 50 \sqrt{ 5 }$

STEP 1: find diagonal $ d1 $

To find diagonal $ d1 $ use formula:

$$ \sin \left( \frac{ \alpha }{ 2 } \right) = \dfrac{ d_2 }{ d_1 } $$

After substituting $ \alpha = 30^o $ and $ d_2 = 25 $ we have:

$$ \sin \left( \frac{ 60^o }{ 2 } \right) = \dfrac{ d_2 }{ } $$ $$ \sin( 30^o ) = \dfrac{ 25 }{ d_1 } $$ $$ \frac{ 1 }{ 2 } = \dfrac{ 25 }{ d_1 } $$ $$ d_1 = \dfrac{ 25 }{ \frac{ 1 }{ 2 } } $$ $$ d_1 = 50 $$

STEP 2: find side $ a $

To find side $ a $ use formula:

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

After substituting $ d_1 = 50 $ and $ d_2 = 25 $ we have:

$$ 50^2 + 25^2 = 4 \cdot a^2 $$ $$ 2500 + 625 = 4 \cdot a^2 $$ $$ 4 \cdot a^2 = 3125 $$ $$ a^2 = \frac{ 3125 }{ 4 } $$ $$ a^2 = \frac{ 3125 }{ 4 } $$ $$ a = \sqrt{ \frac{ 3125 }{ 4 } } $$$$ a = \frac{ 25 \sqrt{ 5}}{ 2 } $$

STEP 3: find perimeter $ P $

To find perimeter $ P $ use formula:

$$ P = 4 \cdot a $$

After substituting $ a = \frac{ 25 \sqrt{ 5}}{ 2 } $ we have:

$$ P = 4 \cdot \frac{ 25 \sqrt{ 5}}{ 2 } $$ $$ P = 50 \sqrt{ 5 } $$

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