若△ABC的三边a,b,c满足条a²+b²+c²+338=10a+24b+26c,试判断△ABC的形状

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若△ABC的三边a,b,c满足条a²+b²+c²+338=10a+24b+26c,试判断△ABC的形状

若△ABC的三边a,b,c满足条a²+b²+c²+338=10a+24b+26c,试判断△ABC的形状
若△ABC的三边a,b,c满足条a²+b²+c²+338=10a+24b+26c,试判断△ABC的形状

若△ABC的三边a,b,c满足条a²+b²+c²+338=10a+24b+26c,试判断△ABC的形状
a²-10a+25+b²-24b+144+c²-26c+169=0
(a-5)²+(b-12)²+(c-13)²=0
a=5,b=12,c=13
是直角三角形

a²+b²+c²+338=10a+24b+26c
a²+b²+c²-10a-24b-26c+338=0
a²-10a+25+b²-24b+144+c²-26c+169=0
(a-5)²+(b-12)²+(c-13)²9=0
a=5
b=12
c=13
a^2+b^2=c^2
所以△ABC为直角△