Question

Problema 3 e urgenttttttttttttt

Answer 1

a)
perimetru=64
perimetru=ab+ac+bc 
dar ab=ac
64=2ab+bc
dar ab=5/6bc
64=2×5/6bc+bc
64=10/6bc+bc aduc la acelasi numitor
64=10bc/6+6bc/6
64=16bc/6⇒bc=64×6 totul supra 16⇒bc=24
ab=5/6 bc⇒ab=5/6×24⇒ab=20
ab=ac⇒ac=20
b) fie ax perpendicular pe bc unde x apartine lui bc 
daca Δabc isoscel⇒xb=xc=bc/2⇒xb=xc=24/12⇒xb/xc=12
in Δaxc aplic teorema piatgora
ax²+xc²=ac²
ax²+12²=20²
ax²=20²-12²
ax²=400-144
ax²=256⇒ax=√256⇒ax=16
arie Δabc=ax×bc totul supra 2
arie Δabc=16×24 totul supra 2
arie Δabc=192
d)R=abc/4A
unde R=raza cerc circumscris
a,b,c=laturile Δ
A=ariaΔ
R=20×20×24/4×192=9600/768=12,5
la c nu stiu sigur si imi e sa nu scriu vreo minune cat capul motanului