在△ABC中,AC=BC=根号2AB⼀2,其周长为4+2倍根号2,求∠A和三角形面积

2025-06-26 17:40:57
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回答1:

AB = (4+2√2)÷(1+√2/2+√2/2) = 2√2 ,
AC = BC = (√2/2)AB = 2 ,
可得:△ABC是等腰直角三角形,
所以,∠A = 45° ;△ABC面积 = (1/2)·AC·BC = 2 .