当球的体积最大时,球与三棱锥D -ABC 的各面相切,设球队半径为R ,则VD -ABC =" VO" -ABC +VO -DAC + VO -DBA + VO -DAB = R(S△ABC + S△DAC + S△DBC + S△DAB).由题设易知AD⊥平面DB
C, 又∵BD
平面DBC,∴AD⊥BD,∴△ABD为直角三角形,∵AB = 4,AD = 3,∴BD = ,∴S△ABC = AD·BD = ×3×= .在△DAB和△DBC中,∵AD = BC,AB = DC,DB = DB,∴△DAB
≌△BCD,故S△DBC = ,VD -ABC =" VA" –DBC = ×3×= ,∴S△ABC = S△ADC = 6,∴R(6 + 6 +
+ ),于是( 4 + )R =
, 解得R =