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