數學問題集(11)---三角形的形狀---I

a^2 + b^2 + c^2 – 19488a – 23384b – 30440c + 463296800 = 0
(a^2 - 19488a + 94945536) + (b^2 - 23384b + 136702864) + (c^2 - 30440c + 231648400) = 0
(a - 9744)^2 + (b - 11692)^2 + (c - 15220)^2 = 0
Since (a - 9744)^2,(b - 11692)^2,(c - 15220)^2 >= 0,
Therefore, a = 9744, b = 11692, c = 15220
Hence, It's an irregular triangle


2010-08-23 19:53:15 補充：
Sorry for my mistakes,
The answer should be right angled triangle, for c is the hypotenuse and right angle is located between side a and b.
Since,
a^2 + b^2 = 9744^2 + 11692^2
= 15220^2
= c^2

2010-08-24 11:52:37 補充：
Since a^2 + b^2 = c^2,
Therefore, it's an right angled triangle, sorry for my mistakes.
其實我尋日已補充-,-""但是不見了-,-""

a = 9744, b = 11692, c = 15220

因為a^2+b^2=c^2

所以是right angle triangle