Question

若三角形ABC的三边a.b.c满足a^2+b^2+c^2+50=6a+8b+10c 试判断△ABC的面积

Answer 1

a^2+b^2+c^2+50=6a+8b+10c 变形就是（a-3）�0�5+（b-4）�0�5+（c-5）�0�5=0 那么a=3 b=4 c=5所以面积为3×4÷2 =6

Answer 2

∵a^2+b^2+c^2+50=6a+8b+10c∴ a^2-6a+9+b^2-8b+16+c^2-10c+25=0∴﹙a-3﹚�0�5+﹙b-4﹚�0�5+﹙c-5﹚�0�5=0∵﹙a-3﹚�0�5≧0,﹙b-4﹚�0�5≧0,﹙c-5﹚�0�5≧0∴a=3,b=4,c=5∴a�0�5+b�0�5=c�0�5=25∠C=90�0�2△ABC的面积 =�0�5×3×4=6