设RT三角形ABC,角C=90度,AD=DC=x,BE=CE=y,AE=3,BD=4
则有:
4x^2+y^2=9
x^2+4y^2=16
两式相加有
5x^2+5y^2=25
得4x^2+4y^2=20即AC^2+BC^2=20
所以AB^2=20,斜边AB=2根号5.
设RT三角形ABC,角C=90度,AD=DC=x,BE=CE=y,AE=3,BD=4
则有:
4x^2+y^2=9
x^2+4y^2=16
两式相加有
5x^2+5y^2=25
得4x^2+4y^2=20即AC^2+BC^2=20
所以AB^2=20,斜边AB=2根号5.