若△ABC的三边a、b、c满足a2+b2+c2+338=10a+24b+26c,则△ABC的面积是(  ) A.338 B.24 C.26 D.30

问题描述:

若△ABC的三边a、b、c满足a2+b2+c2+338=10a+24b+26c,则△ABC的面积是(  )
A. 338
B. 24
C. 26
D. 30

由a2+b2+c2+338=10a+24b+26c,得:(a2-10a+25)+(b2-24b+144)+(c2-26c+169)=0,即:(a-5)2+(b-12)2+(c-13)2=0,a-5=0,b-12=0,c-13=0解得a=5,b=12,c=13,∵52+122=169=132,即a2+b2=c2,∴∠C=90°,...