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In the given figure, if LK = 6\(\sqrt3\) , find MK, ML, KN, MN and the perimeter of \(\square\)MNKL.

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Solution

Solution::

In \(\triangle \space LKM\) , 

\(tan30^∘ ={ML\over LK} \)

\(\Rightarrow ML=6\sqrt { 3 } \times \dfrac { 1 }{ \sqrt { 3 } } =6\)

\(\cos { { 30 }^{ \circ } } =\frac { LK }{ MK }\)

\( \Rightarrow MK=\dfrac { LK }{ \cos { { 30 }^{ \circ } } } =\dfrac { 6\sqrt { 3 } }{ \dfrac { \sqrt { 3 } }{ 2 } } =12\)

 

NK=MK (Side opposite to equal angles )

\(\therefore NK = 12\)

In\(\triangle MKN\)

\(\sin 45^{\circ}=\dfrac{NK}{MN}\)

\(\dfrac{1}{\sqrt2}=\dfrac{12}{MN}\)

\(MN=12\sqrt2\)

Perimeter of ◻MNKL =MN+NK+Lk+ML 

\(perimeter \space = 12\sqrt{2}+12+6\sqrt{3}+6\)

\(= 18+6(2\sqrt2+\sqrt3)\) unit {if cm then cm}

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