設a,b,c表示三角形ABC的三邊長,且滿足a^2+b^2+c^2-ab-bc-ac=0,若三角形abc的周長為18,三角形ABC的面積?
Sol
a^2+b^2+c^2-ab-bc-ac=0
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
(a-b)^2+(a-c)^2+(b-c)^2=0
a=b=c
a+b+c=18
a=6
△ABC=(√3/4)*6^2=9√3
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Verified answer
設a,b,c表示三角形ABC的三邊長,且滿足a^2+b^2+c^2-ab-bc-ac=0,若三角形abc的周長為18,三角形ABC的面積?
Sol
a^2+b^2+c^2-ab-bc-ac=0
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
(a-b)^2+(a-c)^2+(b-c)^2=0
a=b=c
a+b+c=18
a=6
△ABC=(√3/4)*6^2=9√3