不好意思,是有两种情况:
1)顺时针转
∠C = 90°,BC = 4,AC = 3
由勾股定理得到AB = 5
△ABC绕B旋转后,A在BC所在的直线上,
则此时A'B = AB = 5
∴A'C = A'B - BC = 5-4=1
∵AC = 3
根据勾股定理得到AA' = √(9+1) = √10
2)逆时针转
此时A"在BC的左边
A"B = AB = 5
∴A"C = 9
∵AC = 3
∴AA" = √(81+9)= 3√10
不好意思,是有两种情况:
1)顺时针转
∠C = 90°,BC = 4,AC = 3
由勾股定理得到AB = 5
△ABC绕B旋转后,A在BC所在的直线上,
则此时A'B = AB = 5
∴A'C = A'B - BC = 5-4=1
∵AC = 3
根据勾股定理得到AA' = √(9+1) = √10
2)逆时针转
此时A"在BC的左边
A"B = AB = 5
∴A"C = 9
∵AC = 3
∴AA" = √(81+9)= 3√10