BC^2=AB^2-AC^2=21^2-20^2=41
BC=sqrt(41)
Δ ABC ∼ Δ A_(1)B_(1)C_(1)
AC: A_(1)C_(1)=20:5=4:1
AC:A_(1)C_(1)=BC:B_(1)C_(1)=4:1
B_(1)C_(1)=ВС/4=sqrt(41)/4
S_( Δ A_(1)B_(1)C_(1))=(1/2)A_(1)C_(1)*B_(1)C_(1)=(1/2)*5*(sqrt(41)/4)=(5sqrt(41)/8)
б)
СС_(1) ⊥ BC по теореме о трех перпендикулярах
S_(ВB_(1)C_(1)C)=(1/2)(ВС+В_(1)С_(1))*СС_(1)
СС_(1) найдем из прямоугольной трапеции АА_(1)С_(1)С
СС^2_(1)=8^2+(20-5)^2=64+225=289
CC_(1)=17
S_(ВСС_(1)В)=(1/2)((1/4)sqrt(41)+sqrt(41))*17=85sqrt(41)/8
в)
S_(п п)=S_( Δ ABC) + S_ ( Δ A_(1)B_(1)C_(1))+S_(AA_(1)C_(1)C)+S_(ВB_(1)C_(1)C)+S_(AA_(1)В_(1)B)=
=(5sqrt(41)/8) + (1/2)*20*sqrt(41) + ((5+20)*8/2)+ (85sqrt(41)/2)+((21+(21/4))*8/2)=(5sqrt(41)/8)+10sqrt(41)+
=(85sqrt(41)/2)+100+105=[b](115sqrt(41)/2)+205[/b]