Question 779100
You mean, with LEGS of length 6.  This is a Special triangle, 45-45-90 Degreed.  The area is simply, taken one leg as a base, AREA RESULT: {{{(1/2)6*6=highlight(18)}}} square units.


Knowing the area, and leg lengths permit you to find the altitude.  First you can find the longest side, which could act as a base.  Use pythagorean theorem to find this base (which is hypotenuse of the right-isosceles triangle):


{{{b^2=6^2+6^2}}}
{{{b=sqrt(36+36)}}}
{{{b=sqrt(72)}}}
{{{b=sqrt(2*2*3*2*3)}}}
{{{b=6*sqrt(2)}}}


Let A = area of 18
Let b = base of 6*sqrt(2)
Let a = altitude, unknown to be found


Area Formula:  {{{(1/2)ba=A}}}
{{{a=2A/b}}}
Substitute the known values:
{{{highlight(a=2*18/(6*sqrt(2)))}}}
{{{a=6/sqrt(2)}}}
{{{a=(6/2)sqrt(2)}}}
FINAL RESULT: {{{highlight(a=3sqrt(2))}}}