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Solution
AC= $$\sqrt{\ AB^2\ +BC^2}=\ \sqrt{\ 6^2+8^2\ }=10$$
In-radius of right angled triangle = $$\frac{\left(a+b-h\right)}{2}$$
Inradius = $$\frac{\left(6+8-10\right)}{2}=2$$
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