Find the diagonal $ d $ of a rectangle if side $a = \frac{ 3 }{ 2 }$ and perimeter $P = \frac{ 11 }{ 2 }$.

$d = \frac{\sqrt{ 61 }}{ 4 }$

STEP 1: find side $ b $

To find side $ b $ use formula:

$$ P = 2 a + 2 b $$

After substituting $ P = \frac{ 11 }{ 2 } $ and $ a = \frac{ 3 }{ 2 } $ we have:

$$ \frac{ 11 }{ 2 } = 2 \cdot \frac{ 3 }{ 2 } + 2 b $$ $$ \frac{ 11 }{ 2 } = 3 + 2 b $$ $$ 2 b = \frac{ 11 }{ 2 } - 3 $$ $$ 2 b = \frac{ 5 }{ 2 } $$ $$ b = \dfrac{ \frac{ 5 }{ 2 } }{ 2 } $$ $$ b = \frac{ 5 }{ 4 } $$

STEP 2: find diagonal $ d $

To find diagonal $ d $ use Pythagorean Theorem:

$$ a^2 + b^2 = d^2 $$

After substituting $ a = \frac{ 3 }{ 2 } $ and $ b = \frac{ 5 }{ 4 } $ we have:

$$ \left(\frac{ 3 }{ 2 }\right)^2 + \left(\frac{ 5 }{ 4 }\right)^2 = d^2 $$ $$ \frac{ 9 }{ 4 } + \frac{ 25 }{ 16 } = d^2 $$ $$ d^2 = \frac{ 61 }{ 16 } $$ $$ d = \sqrt{ \frac{ 61 }{ 16 } } $$$$ d = \frac{\sqrt{ 61 }}{ 4 } $$

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