矩形ABCD中AB＝4,BC＝3,沿AC将矩形ABCD折起,使面BAC⊥面DAC,求四面体A—BCD的外接球...

题目

矩形ABCD中AB＝4,BC＝3,沿AC将矩形ABCD折起,使面BAC⊥面DAC,求四面体A—BCD的外接球...
矩形ABCD中AB＝4,BC＝3,沿AC将矩形ABCD折起,使面BAC⊥面DAC,求四面体A—BCD的外接球的体积.

答案

设矩形ABCD中,AC交BD与O,则OA=OB=OC=OD
故,O为四面体A-BCD的外接球的球心.
故R=1/2AC
解△ABC得,AC==√(AB^2+BC^2)=5
R=2.5
V=4/3(πR^3)=125π/6