设直四棱柱ABCD-A1B1C1D1,底四边形ABCD是菱形,二条体对角线AC1,BD1,AC1=9,BD1=15,高CC1=5,AC和BD交于O点,
分别连结AC、BD、A1C1和B1D1,
〈ACC1=90度,
根据勾股定理,
AC1^2=AC^2+CC1^2,
AC=√(9^2-5^2)=2√14,
BD^2=BD1^2-DD1^2,
BD=√(15^2-5^2)=10√2,
AO=AC/2=√14,
BO=BD/2=5√2,
AB^2=AO^2+BO^2,
AB=8,
V=(AC*BD/2)*AA1=[(2√14)*(10√2)/2]*5=100√7.
表面积S=AB*AA1*4+（AC*BD/2）*2
=160+40√7.