Ответ:

1) tgBAC=3/4=BC/AC => BC=3x, AC=4xAB=(AC^2+BC^2)^0.5=5x

 

2) SΔABC = 0.5*AC*BC=6x^2SΔABC = 0.5*AB*CP=2.5x*CP6x^2=2.5x*CP; CP = 2.4×3)BP=(BC^2-CP^2)^0.5=1.8×4) BCD: r=0.5(CP+BP-CB)=0.5(1.8x-2,4x-3x)=8; x=40/35) ABC: r=0.5(AC+BC-AB)=0.5*40/3(4+3-5)=40/3=13⅓Ответ: 13⅓