Question 600333
Hypotenuse = 6m

one leg = {{{2sqrt(5)}}}

Let other leg be x

Apply Pythagoras theorem 

{{{(6)^2= x^2+ (2sqrt(5))^2}}}


{{{6^2 -(2sqrt(5))^2= x^2}}}

{{{36-20= x^2}}}


16=x^2

x= +/- 4

Ignore negative value

other leg = 4 m