What is the radius?

Let the circle be tangent to our triangle's sides AB,BC,CA at points P,Q,R respectively ,then recall that tangents from a point to a circle are always equidistant...

Thus,AP=AR=8-r also CQ=CR=6-r

AR + CR = AC = 10

(8-r)+(6-r)=10

2r = 4

r=2

Its simple just remember; {(AB+BC)-AC}/2----------------(1)

According to the Pythagoras theorem;

AB^{2}=BC^{2}=AC^{2}

8^{2}+6^{2}=AC^{2}

64+36=AC^{2}

100=AC^{2}

10=AC---------------(2)

PUT THE VALUE OF (2) INTO (1);

{(8+6)}-10}/2

4/2

=2

Thus radius(r)=2

Thank you.