Question

如果三角形ABC的三边长分别为a,b,c.且满足a^2+b^2+c^2+50=6a+8b+10c,试判断三角形的形状

Answer 1

解：a�0�5+b�0�5+c�0�5+50=6a+8b+10ca�0�5-6a+9+b�0�5-8b+16+c�0�5-10c+25=0（a-3）�0�5+（b-4）�0�5+（c-5）�0�5=0所以a=3，b=4，c=5a�0�5=3�0�5=9b�0�5=4�0�5=16c�0�5=5�0�5=25所以a�0�5+b�0�5=c�0�5=25所以三角形ABC是直角三角形

Answer 2

直角三角形a�0�5+b�0�5+c�0�5+50=6a+8b+10ca�0�5-6a+9+b�0�5-8b+16+c�0�5-10c+25=0（a-3）�0�5+（b-4）�0�5+（c-5）�0�5=0所以a=3，b=4，c=5所以a�0�5+b�0�5=c�0�5=25所以三角形ABC是直角三角形

Answer 3

a2+b2+c2+50=6a+8b+10c
a^2-6a+9+b^2-8b+16+c^2-10b+25=0
(a-3)^2+(b-4)^2+(c-5)^2=0
所以(a-3)^2=0，(b-4)^2=0，(c-5)^2=0 a=3,b=4,c=5
勾三股四弦五
3^3+4^2=5^2
三角形为直角三角形，

Answer 4

由a^2+b^2+c^2+50=6a+8b+10c得（a-3)^2+(b-4）^2+(c-5)^2=0所以a=3 b=4 c=5直角三角形