如图，圆台的轴截面为ABCD，截得此圆台的圆锥的轴截面为OAB，圆台上底半径为FC=6，下底半径为EB=12，截面半径为MN，
则由题意可得，可令CN=2x，BN=x．
由直角三角形OFC与直角三角形OEB相似可得
| FC |
| EB |
| OC |
| OB |
| 6 |
| 12 |
| OC |
| OC+2x+x |
∴OC=3x．
再由△PMN与△OEB相似可得
| MN |
| EB |
| ON |
| OB |
| MN |
| 12 |
| 3x+2x |
| 3x+2x+x |
故截面的面积为 π MN2=100π，
故答案为 100π．
如图，圆台的轴截面为ABCD，截得此圆台的圆锥的轴截面为OAB，| FC |
| EB |
| OC |
| OB |
| 6 |
| 12 |
| OC |
| OC+2x+x |
| MN |
| EB |
| ON |
| OB |
| MN |
| 12 |
| 3x+2x |
| 3x+2x+x |