The perimeter of a triangle right angled at  $\text{'}C\text{'}$ is 70, and the in-radius is 6, then $|a-b|=$

$\Delta =rs$

$=\frac{70}{2}\times 6$

$=35\times 6$

$=210$

$\frac{1}{2}ab=210$

$ab=420$

$a+b+c=70$

$a+b=70-c$

${(a+b)}^2={(70-c)}^2$

$a^2+b^2+2ab=4900-140c+c^2$

$c=\frac{4900-2\times 420}{140}=29[\therefore a^2+b^2=c^2]$

${(a-b)}^2=a^2+b^2-2ab$

$=c^2-2ab$

$=841-840$

$\therefore a-b=1$