题目内容
(2012•桂林一模)半径为4的球面上有A,B,C,D四点,且满足AB⊥AC,AC⊥AD,AD⊥AB,则S△ABC+S△ACD+S△ADB的最大值为(S为三角形的面积)
32
32
.分析:设AB=a,AC=b,AD=c,根据AB⊥AC,AC⊥AD,AD⊥AB,可得a2+b2+c2=4R2=64,而S△ABC+S△ACD+S△ADB=
(ab+ac+bc),利用基本不等式,即可求得最大值为.
1 |
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解答:解:设AB=a,AC=b,AD=c,
∵AB⊥AC,AC⊥AD,AD⊥AB,∴a2+b2+c2=4R2=64
∴S△ABC+S△ACD+S△ADB=
(ab+ac+bc)≤
(a2+b2+c2)=32
∴S△ABC+S△ACD+S△ADB的最大值为32
故答案为:32.
∵AB⊥AC,AC⊥AD,AD⊥AB,∴a2+b2+c2=4R2=64
∴S△ABC+S△ACD+S△ADB=
1 |
2 |
1 |
2 |
∴S△ABC+S△ACD+S△ADB的最大值为32
故答案为:32.
点评:本题考查求内接几何体,考查基本不等式的运用,属于基础题.
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