The above figure represents the tetrahedron as described in the problem. Here three mutually perpendicular edges of the tetrahedron with lengths 3 cm, 4 cm, and 5 cm are placed along positive direction of X-axis,Y-axis and Z-axis respectively positioning vertex at origin #O#
Let us consider a triangle #PQR# parallel to the triangular base #ABC# at the height #z#cm from base.
Now #DeltaOAT and Delta PQT# are similar
So #(PQ)/(OA)=(TP)/(OT)#
#=>(PQ)/3=(5-z)/5#
#=>PQ=3xx(5-z)/5#
Again #DeltaOBT and Delta PRT# are similar
So #(PR)/(OB)=(TP)/(OT)#
#=>(PR)/4=(5-z)/5#
#=>PR=4xx(5-z)/5#
Now in #DeltaRPQ #, #/_RPQ=90^@#
So area of #DeltaRPQ=a=1/2xxPQxxPR=1/2xx3xx4 ((5-z)/5)^2#
#=6xx ((5-z)/5)^2#
Considering an infinitesimally small thickness #dz# over #DeltaRPQ # and finding the value of following integral we can get the volume of the given tetrahedron
So the volume of the tetrahedron
#V=int_0^5adz=int_0^5 6xx ((5-z)/5)^2dz#
#=6/25int_0^5 (5-z)^2dz#
#=-6/25xx[ (5-z)^3/3]_0^5#
#=-6/25xx[ (5-5)^3/3-(5-0)^3/3]#
#=6/25xx 5^3/3=10#cc
