Question

已知,直角三角形的外接圆半径为5,内切圆半径为1 则三角形周长为

Answer 1

RT△ABC的外接圆半径R=5,∠C=90°,内切圆r=1

设内切圆与斜边AB切于D,AC切于E,BC切于F,则

CE=CF=r=1,BF=BD,AD=AE

AB=AD+BD=2R

RT△ABC的周长

=AB+AC+BC=AD+BD+AE+CE+CF+BF

=CE+CF+AD+BD+AE+BF

=1+1+2*2R

=2+4*5

=22