se ia trapezul dreptunghic OBB'O'
ducem din B' perpendiculara pe OB si notam cu N
NB=OB-O'B'=11-5=6
ΔB'NB, mas<N=90 ⇒BB'²= NB²+B'N²= 6²+8²=36+64=100 ⇒BB'=10
g=BB'=10
Al=πg (R+r)= 10π( 11+5)= 10*16π= 160π
V=πh (R²+r²+Rr)/3= 8π (11²+5²+5*11)/3= 8π (121+25+55)/3= 8π *201/3=8* 67π=536π
