Question

1,三角形ABC的三边长a,b,c满足b+c=8,bc=a的平方减12a+52,则三角形ABC的周长为多少?

Answer 1

+c≥2√bc

bc≤[(b+c)/2]^2=4^2=16

bc=a^2-12a+52≤16

a^2-12a+36≤0

(a-6)^2≤0

a-6=0

a=6

C=a+b+c=6+8=14