Question

Ma puteți ajuta la acest exercitiu? Va rog frumos!

Answer 1

p=(AB+BC+AC)/2=(50+50+60)/2=160/2=80 u

r=√[(p-AB)(p-BC)(p-AC)/p]=√[(80-50)(80-50)(80-60)/80]=√[(30*30*20)/80]=√(18000/80)=√225=15 u=>r=15 u

A=p*r=80*15=1200 u²

A=AB*BC*AC/4R=>4R=AB*BC*AC/A=>R=AB*BC*AC/4A=50*50*60/(4*1200)=150000/4800=1500/48=375/12=125/4 u

R=125/4 u

Answer 2

Explicație pas cu pas:

R-raza cercului circumscris

r-raza cercului inscris

1.In acest triunghi isoscel ducem inaltimea AD, AD⊥BC;

AD=√(AB²-(BC/2)²)=√(50²-30²)=√(50-30)(50+30)=√(20*80)=√1600=40(u)

sinB=AD/AB=40/50=4/5

Aplicam Teorema Sinusurilor:

AC/sinB=2*R⇒50/4/5=2*R⇒250/4=2*R⇒R=250/8⇒R=125/4(u)

2.

r=4*R*sinA/2*sinB/2*sinC/2

sinA/2=3/5

sinB/2=√(1-cosB)/2

cosB=√(1-sin²B)=√(1-16/25)=√(9/25)=3/5

sinC/2=sinB/2=√(1-3/5)/2=√(2/5/2)=√5/5

r=4*125/4*3/5*(√5/5)²=25*3*5/25=15(u)

Raspuns: R=125/4(u) r=15(u)

Bafta!