Question

解三角形\在三角形ABC中,三边a.b.c满足a+b+c=3/2根号2,a平方+b平方+c平方=3/2,判断三角形形状

Answer 1

∵a+b+c=3√2/2

∴(a+b+c)²=9/2=3*(3/2)=3(a²+b²+c²)

a²+b²+c²+2ab+2bc+2ac=3a²+3b²+3c²

2a²+2b²+2c²-2ab-2bc-2ac=0

(a-b)²+(b-c)²+(a-c)²=0

(a-c)²=0 即a=c

(b-c)²=0 即b=c

(a-b)²=0 即a=b

∴△ABC是等边三角形