Step-by-step explanation:
[tex] \triangle [/tex] JKL is a 45° - 45° - 90° triangle.
Side opposite to 45° is [tex] \frac{1}{\sqrt 2}[/tex] times of hypotenuse.
[tex] \therefore \: JK = \frac{1}{ \sqrt{2} } \times JL \\ \\ \therefore \: 6 \sqrt{3} = \frac{1}{ \sqrt{2} } \times JL \\ \\ \therefore \: 6 \sqrt{3} \times \sqrt{2} = JL \\ \\ \therefore \: 6 \sqrt{3 \times 2} = JL \\ \\ \huge \red{ \boxed{\therefore \:JL = 6 \sqrt{6} \: units.}} \\ \\ \because \angle \: J =\angle \: L = 45 \degree \\ \\ \purple{ \boxed{ \therefore \: KL = JK = 6 \sqrt{3} \: units}}[/tex]
