Question 954845
With height 4in and legs 5in, a 3-4-5 triangle is formed,
with the third side (3 in) being how much longer one base is
than the other on each end. So one base is 6 inches longer.
a=shorter base; b=longer base=a+6in; h=height=4in; A=Area=77 sq in
{{{A=((a+b)/2)h}}} Substitute for b
{{{A=((a+(a+6in))/2)h}}}
{{{77in^2=((2a+6in)/2)4in}}}  
{{{77in^2=(a+3in)4in}}}
{{{77in^2=4in(a)+12in^2}}} Subtract {{{12in^2}}} from each side
{{{65in^2=4in(a)}}} Divide each side by 4in
{{{16.25in=a}}} ANSWER 1: The shorter base is 16.25 inches.
{{{a+6in=16.25+6in=22.25in}}} ANSWER 2: The longer base is 22.25 inches.
CHECK:
{{{A=((a+b)/2)h}}}
{{{77in^2=((16.25in+22.25)/2)4in}}}
{{{77in^2=((38.5in/2)4in)}}}
{{{77in^2=(19.25in)4in}}}
{{{77in^2=77in^2}}}