Question

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

Answer 1

a^2-10a+25+b^2-24b+144+c^2-26c+169=25+144+169

(a-5)^2+(b-12)^2+(c-13)^2=25+144+169

a=10

b=24

c=26

10^2+24^2=26^2

所以符合勾股定理

所以是直角三角形