Re: [中學] 垂心
※ 引述《justin0602 (justin)》之銘言:
: 已知三角形ABC
: AB=4
: BC=6
: AC=2sqrt(7) 兩倍根號七
: 求AH的長度 2 2 2
__\ __\ __\ __\ __\ __\ 4 +(2√7) -6
AH ˙AB = AH ˙AC =AB ˙AC = ------------ = 4
2
__\ __\ __\
設 AH = a AB + b AC
__\ __\ __\ 2 __\ __\
AH ˙AB = a |AB | + b AB ˙AC => 4 = 16a + 4b => 1 = 4a + b
__\ __\ __\ __\ __\ 2
AH ˙AC = a AB ˙AC + b |AC | => 4 = 4a + 28b =>1 = a + 7b
2 1 __\ 2 4 4 1 108
a = ---, b = --- => |AH | = ----×16+----×4+----×28 = -----
9 9 81 81 81 81
__ 6√3 2√3
=> AH = ---- = -----
9 3
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