$解：作OC⊥AB于O，則OC為兩個圓錐共同的底面$
$的半徑，如圖所示$
$則AB= \sqrt{A{C}^2+B{C}^2}= \sqrt{{3}^2+{4}^2}$
$=5(\ \mathrm {cm})$
$因為AB·OC=AC·BC$
$所以O(shè)C= \frac {12}{5}\ \mathrm {cm}$
$以AC為母線的圓錐側(cè)面積= \frac {1}{2}×2π× \frac {12}{5}×3= \frac {36}{5}π(\ \mathrm {cm}2)$
$以BC為母線的圓錐側(cè)面積= \frac {1}{2}×2π× \frac {12}{5}×4= \frac {48}{5}π(\ \mathrm {cm}2)$
$所以表面積為 \frac {36}{5}π+ \frac {48}{5}π= \frac {84}{5}π\(zhòng) \mathrm {cm}2$
