Question

exercitiu 2 de sus va rog

Answer 1

a. V=Abxh=20x20x10=4000 cm²
b.At=Al+2Ab
Al=Pbxh=4x20x10=800 cm²
Ab=20x20=400 cm²
At=800+2x400=1600 cm²
1600........100%
     x..........110%
x=1600x110/100=1760 cm²
c. M este situat pe BO astfel incat distanta CM sa fie minima⇒CM⊥OB
ΔHGF Δdreptunghic, t Pitagora HF²=HG²+GF²=400+400=800, HF=20√2
OF=HF/2=10√2
ΔOBF dreptunghic OB²=OF²+FB², OB=10√3
in ΔOBC isoscel (OB=OC) ducem ON⊥BC, inaltime, mediana ,mediatoare, BN=NC=10 cm
ON²=OB²-BN², ON=10√2
Aria (OBC)=OMxBC/2=10√2x20/2=100√2
Aria(OBC)=MCxOB/2, MC=2Aria/OB=2x100√2/10√3=20√2/√3=20√6/3