$解：連接?OE，?設(shè)扇形?ODF ?的半徑為?r?$
$∵點(diǎn)?E?為圓的切點(diǎn)$
$∴?∠OEB=90°?$
$∴?∠OEB=∠ACB?$
$∵?∠B?為公共角$
$∴?△OEB∽△ACB?$
$∴?\frac {OE}{AC}=\frac {OB}{AB}?$
$∵?AC=6\ \mathrm {cm}，??BC=8\ \mathrm {cm}?$
$∴?AB=10\ \mathrm {cm}?$
$∴?OB=\frac 53r，??AO=10-\frac 53r?$
$∵?∠AOF=∠ACB，??∠A?為公共角$
$∴?△AOF∽△ACB?$
$∴?\frac {AO}{AC}=\frac {OF}{BC}，??\frac {10-\frac 53r}6=\frac {r}8?$
$∴?r=\frac {120}{29}\ \mathrm {cm}?$
