题目内容
边长为a,b,c的三角形有面积公式(海伦公式):S=
,其中s为半周,即s=
(a+b+c).若△ABC的三边a,b,c满足a2+b2+c2=16,a4+b4+c4=96.则S△ABC=______.
| s(s-a)(s-b)(s-c) |
| 1 |
| 2 |
∵a2+b2+c2=16,a4+b4+c4=96,
又∵(a2+b2+c2)2=a4+b4+c4+2a2b2+2b2c2+2a2c2,
即162=96+(2a2b2+2b2c2+2a2c2),
∴2a2b2+2b2c2+2a2c2=160,
∵s=
(a+b+c),
∴s(s-a)(s-b)(s-c)=
(a+b+c)•
(b+c-a)•
(a+c-b)•
(a+b-c)=
[(2a2b2+2b2c2+2a2c2)-(a4+b4+c4)]=
×(160-96)=4,
∴S△ABC=
=
=2.
故答案为:2.
又∵(a2+b2+c2)2=a4+b4+c4+2a2b2+2b2c2+2a2c2,
即162=96+(2a2b2+2b2c2+2a2c2),
∴2a2b2+2b2c2+2a2c2=160,
∵s=
| 1 |
| 2 |
∴s(s-a)(s-b)(s-c)=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 16 |
| 1 |
| 16 |
∴S△ABC=
| s(s-a)(s-b)(s-c) |
| 4 |
故答案为:2.
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