Question 358907
{{{H^2=(H-4)^2+(H-8)^2}}}
{{{H^2=H^2-8H+16+H^2-16H+64}}}
{{{H^2-24H+80=0}}}
{{{(H-4)(H-20)=0}}}
Two solutions:
{{{H-4=0}}}
{{{H=4}}} 
However, one leg would be zero length, the other would be negative.
No solution.
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.
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{{{H-20=0}}}
{{{highlight(H=20)}}}cm
Then the two legs are {{{16}}}cm and {{{12}}}cm