若RT三角形的内切圆与斜边AB相切于D,且AD=1,BD=2,则S ABC=?

问题描述:

若RT三角形的内切圆与斜边AB相切于D,且AD=1,BD=2,则S ABC=?

设内切圆半径为 r ;已知,AD = 1 ,BD = 2 ,可得:BC = 2+r ,AC = 1+r ,AB = 1+2 = 3 ,所以,S△ABC = ½(BC+AC+AB)r = r²+3r ;由勾股定理可得:BC²+AC² = AB² ,即有:(2+r)²+(1+r)...