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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::

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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