Question 898895
x for leg
h for hypotenuse
{{{x=h-3}}}, if the description about this is understood correctly.


Expect because of right, isosceles, that {{{x^2+x^2=h^2}}} applying Pythagorean Theorem.
{{{2x^2=h^2}}}.


Perimeter is given, 16.
{{{x+x+h=16}}}
{{{2x+h=16}}}


A system of two equations is now possible.
{{{system(2x^2=h^2,2x+h=16)}}}.


{{{h=16-2x}}}
{{{h=2(8-x)}}}
substitute into the Pythagorean simplified equation;
{{{2x^2=(2(8-x))^2}}}
{{{2x^2=4(8-x)^2}}}
{{{x^2=2(8-x)^2}}}
{{{x^2=2(64-18x+x^2)}}}
{{{x^2=128-36x+x^2}}}
{{{0=128-36x}}}
{{{36x=128}}}
{{{x=128/36}}}
{{{highlight(x=32/9)}}}


Use the perimeter fact to find h, the hypotenuse.
{{{h=16-2x}}}
{{{h=16-2(32/9)}}}
{{{h=(16*9-2*32)/9}}}
{{{h=(144-64)/9}}}
{{{highlight(h=80/9)}}}


Is h-x=3?
{{{80/9-32/9}}}
{{{(80-32)/9}}}
{{{48/9}}}
{{{highlight_green(h-x=5&1/3<>3)}}}-----That part of the problem description is WRONG.