Question

在RT△ABC中,∠ACB=90°,CD垂直AB,AB=13,CD=6,(AC+BC)的平方?

Answer 1

由勾股定理,AC^2+BC^2=AB^2=13^2=169,

由三角形面积,AC*BC/2=AB*CD/2,AC*BC=78,

(AC+BC)^2

=AC^2+BC^2+2AC*BC

=169+2*78

=169+156

=325