Question 955553
{{{a+c+b=16}}}; {{{a=c}}};
{{{2a+b=16}}}


{{{(b/2)^2+h^2=a^2}}}
Given h=4;
{{{(b/2)^2+16=a^2}}}.


You may be able to ---- solve the system {{{highlight_green(2a+b=16,(b/2)^2+16=a^2)}}}, <s>and not need to use the given area.  The system is two equations in two unknowns.</s>


The area equation may be 
{{{highlight_green((1/2)*b*4=12)}}}, and this may make finding c easier to do.  Maybe not.
.....this area means, {{{b=6}}} and you already know {{{a+b+c=16}}},
{{{b+a+c=16}}}
{{{b+2c=16}}}
{{{6+2c=16}}}
{{{2c=10}}}
{{{highlight(c=5)}}}