$解：作OC⊥AB于O，则OC为两个圆锥共同的底面$
$的半径，如图所示$
$则AB= \sqrt{A{C}^2+B{C}^2}= \sqrt{{3}^2+{4}^2}$
$=5(\ \mathrm {cm})$
$因为AB·OC=AC·BC$
$所以OC= \frac {12}{5}\ \mathrm {cm}$
$以AC为母线的圆锥侧面积= \frac {1}{2}×2π× \frac {12}{5}×3= \frac {36}{5}π(\ \mathrm {cm}²)$
$以BC为母线的圆锥侧面积= \frac {1}{2}×2π× \frac {12}{5}×4= \frac {48}{5}π(\ \mathrm {cm}²)$
$所以表面积为 \frac {36}{5}π+ \frac {48}{5}π= \frac {84}{5}π\ \mathrm {cm}²$
