Question 113059
You really can't solve this problem the way you posed it because there are an infinite number of different right triangles with a hypotenuse = {{{6*sqrt(3)}}}.  Now if you had specified this was an isoceles right triangle, meaning both of the legs were of equal length or if you had specified any other proportional relationship between the length of the legs, then the problem becomes solveable.


I'm going to assume you meant an isoceles right triangle and solve this.


Pythagoras tells us that for any right triangle of sides a, b, and hypotenuse c, the following relationship is true:


{{{a^2+b^2=c^2}}}


We are given that {{{c=6*sqrt(3)}}}, so {{{c^2=36*3=108}}}.


So now we can write:
{{{a^2+b^2=108}}}


But, since we are assuming that this is an isoceles right triangle, a = b, so we can now write:
{{{a^2+a^2=108}}}


So now, let's simplify and solve:
{{{2a^2=108}}}, combine like terms
{{{a^2=54}}}, divide both sides by 2
{{{a=sqrt(54)=sqrt(6*9)=3*sqrt(6)}}}, take the square root of both sides. Done.


Except we should check our answer:
{{{(3*sqrt(6))^2+(3*sqrt(6))^2=(6*sqrt(3))^2}}}
{{{(9*6)+(9*6)=108}}}
{{{54+54=108}}}, Checks.


Now what if one leg of the triangle had been specified to be three times longer than the other?


Then, instead of a =  b, we would say that 3a = b,
{{{a^2+b^2=108}}}
{{{a^2+(3a)^2=108}}}
{{{a^2+9a^2=108}}}
{{{10a^2=108}}}
{{{a^2=10.8}}}
{{{a=sqrt(10.8)}}} and {{{b=3*sqrt(10.8)}}}


Hope this helps,
John